R. Bartle, D. Sherbert - Instructors Manual - Introduction to Real Analysis-John Wiley & Sons (2010).pdf

INSTRUCTOR’S MANUAL TO ACCOMPANY
INTRODUCTION TO REAL
ANALYSIS
Fourth Edition
Robert G. Bartle
Eastern Michigan University
Donald R. Sherbert
University of Illinois
JOHN WILEY & SONS, INC.
New York
•Chichester
•Weinheim
•Brisbane
•Singapore
•Toronto

Copyrightc2000, 2010 by John Wiley & Sons, Inc.
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ISBN 978-0-471-44799-3

PREFACE
This manual is oﬀered as an aid in using the fourth edition ofIntroduction to Real
Analysisas a text. Both of us have frequently taught courses from the earlier
editions of the text and we share here our experience and thoughts as to how to
use the book. We hope our comments will be useful.
We also provide partial solutions for almost all of the exercises in the book.
Complete solutions arealmost neverpresented here, but we hope that enough is
given so that a complete solution is within reach. Of course, there is more than
one correct way to attack a problem, and you may ﬁnd better proofs for some of
these exercises.
We also repeat the graphs that were given in the manual for the previous
editions, which were prepared for us by Professor Horacio Porta, whom we wish
to thank again.
Robert G. Bartle November 20, 2010
Donald R. Sherbert

CONTENTS
Chapter 1 Preliminaries.....................................................1
Chapter 2 The Real Numbers...............................................7
Chapter 3 Sequences.......................................................17
Chapter 4 Limits...........................................................28
Chapter 5 Continuous Functions...........................................33
Chapter 6 Diﬀerentiation...................................................43
Chapter 7 The Riemann Integral...........................................51
Chapter 8 Sequences of Functions..........................................61
Chapter 9 Inﬁnite Series...................................................68
Chapter 10 The Generalized Riemann Integral.............................77
Chapter 11 A Glimpse into Topology.......................................88
Selected Graphs.............................................................95

CHAPTER 1
PRELIMINARIES
We suggest that this chapter be treated as review and covered quickly, without
detailed classroom discussion. For one reason, many of these ideas will be already
familiar to the students — at least informally. Further, we believe that, in practice,
those notions of importance are best learned in the arena of real analysis, where
their use and signiﬁcance are more apparent. Dwelling on the formal aspect of
sets and functions does not contribute very greatly to the students’ understanding
of real analysis.
If the students have already studied abstract algebra, number theory or com-
binatorics, they should be familiar with the use of mathematical induction. If not,
then some time should be spent on mathematical induction.
The third section deals with ﬁnite, inﬁnite and countable sets. These notions
are important and should be brieﬂy introduced. However, we believe that it is
not necessary to go into the proofs of these results at this time.
Section 1.1
Students are usually familiar with the notations and operations of set algebra,
so that a brief review is quite adequate. One item that should be mentioned is
that two setsAandBare often proved to be equal by showing that: (i) ifx∈A,
thenx∈B, and (ii) ifx∈B, thenx∈A. This type of element-wise argument is
very common in real analysis, since manipulations with set identities is often not
suitable when the sets are complicated.
Students are often not familiar with the notions of functions that are injective
(= one-one) or surjective (= onto).
Sample Assignment: Exercises 1, 3, 9, 14, 15, 20.
Partial Solutions:
1. (a)B∩C={5,11,17,23,...}={6k−1:k∈N},A∩(B∩C)={5,11,17}
(b) (A∩B)\C={2,8,14,20}
(c) (A∩C)\B={3,7,9,13,15,19}
2. The sets are equal to (a)A, (b)A∩B,(c) the empty set.
3. IfA⊆B, thenx∈Aimpliesx∈B, whencex∈A∩B, so thatA⊆A∩B⊆A.
Thus, ifA⊆B, thenA=A∩B.
Conversely, ifA=A∩B, thenx∈Aimpliesx∈A∩B, whencex∈B.
Thus ifA=A∩B, thenA⊆B.
4. Ifxis inA\(B∩C), thenxis inAbutx/∈B∩C, so thatx∈Aandxis
either not inBor not in C. Therefore eitherx∈A\Borx∈A\C, which
implies thatx∈(A\B)∪(A\C). ThusA\(B∩C)⊆(A\B)∪(A\C).
1

2 Bartle and Sherbert
Conversely, ifxis in (A\B)∪(A\C), thenx∈A\Borx∈A\C.Thus
x∈Aand eitherx/∈Borx/∈C, which implies thatx∈Abutx/∈B∩C,
so thatx∈A\(B∩C). Thus (A\B)∪(A\C)⊆A\(B∩C).
Since the setsA\(B∩C) and (A\B)∪(A\C) contain the same elements,
they are equal.
5. (a) Ifx∈A∩(B∪C),thenx∈Aandx∈B∪C. Hence we either have
(i)x∈Aandx∈B, or we have (ii)x∈Aandx∈C. Therefore, either
x∈A∩Borx∈A∩C, so thatx∈(A∩B)∪(A∩C). This shows that
A∩(B∪C) is a subset of (A∩B)∪(A∩C).
Conversely, letybe an element of (A∩B)∪(A∩C). Then either (j)y∈
A∩B, or (jj)y∈A∩C. It follows thaty∈Aand eithery∈Bory∈C.
Therefore,y∈Aandy∈B∪C, so thaty∈A∩(B∪C). Hence (A∩B)∪
(A∩C) is a subset ofA∩(B∪C).
In view of Deﬁnition 1.1.1, we conclude that the setsA∩(B∪C) and
(A∩B)∪(A∩C) are equal.
(b) Similar to (a).
6. The setDis the union of{x:x∈Aandx/∈B}and{x:x/∈Aandx∈B}.
7. HereA
n={n+1,2(n+1),...}.
(a)A
1={2,4,6,8,...},A 2={3,6,9,12,...},A 1∩A2={6,12,18,24,...}=
{6k:k∈N}=A
5.
(b)
ﬀ
A
n=N\{1}, because ifn>1, thenn∈A n−1; moreover 1/∈A n.
Also
ﬃ
A
n=∅, becausen/∈A nfor anyn∈N.
8. (a) The graph consists of four horizontal line segments.
(b) The graph consists of three vertical line segments.
9. No. For example, both (0, 1) and (0,−1) belong toC.
10. (a)f(E)={1/x
2
:1≤x≤2}={y:
1
4
≤y≤1}=[
1
4
,1].
(b)f
−1
(G)={x:1≤1/x
2
≤4}={x:
1
4
≤x
2
≤1}=[−1,−
1
2
]∪[
1
2
,1].
11. (a)f(E)={x+2:0≤x≤1}=[2,3], soh(E)=g(f(E)) =g([2,3]) =
{y
2
:2≤y≤3}=[4,9].
(b)g
−1
(G)={y:0≤y
2
≤4}=[−2,2], soh
−1
(G)=f
−1
(g
−1
(G)) =
f
−1
([−2,2]) ={x:−2≤x+2≤2}=[−4,0].
12. If 0 is removed fromEandF, then their intersection is empty, but the
intersection of the images underfis{y:0<y≤1}.
13.E\F={x:−1≤x<0},f(E)\f(F) is empty, andf(E\F)=
{y:0<y≤1}.
14. Ify∈f(E∩F), then there existsx∈E∩Fsuch thaty=f(x). Sincex∈E
impliesy∈f(E), andx∈Fimpliesy∈f(F), we havey∈f(E)∩f(F). This
provesf(E∩F)⊆f(E)∩f(F).
15. Ifx∈f
−1
(G)∩f
−1
(H), thenx∈f
−1
(G) andx∈f
−1
(H), so thatf(x)∈G
andf(x)∈H. Thenf(x)∈G∩H,and hencex∈f
−1
(G∩H). This shows

Chapter 1 — Preliminaries 3
thatf
−1
(G)∩f
−1
(H)⊆f
−1
(G∩H). The opposite inclusion is shown in
Example 1.1.8(b). The proof for unions is similar.
16. Iff(a)=f(b), thena/
√
a
2
+1=b/
√
b
2
+ 1, from which it follows thata
2
=b
2
.
Sinceaandbmust have the same sign, we geta=b, and hencefis injective.
If−1<y<1, thenx:=y/
Σ
1−y
2
satisﬁesf(x)=y(why?), so thatftakesR
onto the set{y:−1<y<1}.Ifx>0, thenx=
√
x
2
<
√
x
2
+ 1, so it follows
thatf(x)∈{y:0<y<1}.
17. One bijection is the familiar linear function that mapsato 0 andbto 1,
namely,f(x):=(x−a)/(b−a).Show that this function works.
18. (a) Letf(x)=2x,g(x)=3x.
(b) Letf(x)=x
2
,g(x)=x,h(x) = 1. (Many examples are possible.)
19. (a) Ifx∈f
−1
(f(E)), thenf(x)∈f(E), so that there existsx 1∈Esuch
thatf(x
1)=f(x). Iffis injective, thenx 1=x, whencex∈E. Therefore,
f
−1
(f(E))⊆E. SinceE⊆f
−1
(f(E)) holds for anyf, we have set equality
whenfis injective. See Example 1.1.8(a) for an example.
(b) Ify∈Handfis surjective, then there existsx∈Asuch thatf(x)=y.
Thenx∈f
−1
(H) so thaty∈f(f
−1
(H)). ThereforeH⊆f(f
−1
(H)). Since
f(f
−1
(H))⊆Hfor anyf, we have set equality whenfis surjective. See
Example 1.1.8(a) for an example.
20. (a) Sincey=f(x) if and only ifx=f
−1
(y),it follows thatf
−1
(f(x)) =xand
f(f
−1
(y)) =y.
(b) Sincefis injective, thenf
−1
is injective onR(f). And sincefis surjec-
tive, thenf
−1
is deﬁned onR(f)=B.
21. Ifg(f(x
1)) =g(f(x 2)),thenf(x 1)=f(x 2), so thatx 1=x2, which implies that
g◦fis injective. Ifw∈C, there existsy∈Bsuch thatg(y)=w, and there
existsx∈Asuch thatf(x)=y. Theng(f(x)) =w, so thatg◦fis surjective.
Thusg◦fis a bijection.
22. (a) Iff(x
1)=f(x 2), theng(f(x 1)) =g(f(x 2)), which impliesx 1=x2, since
g◦fis injective. Thusfis injective.
(b) Givenw∈C, sinceg◦fis surjective, there existsx∈Asuch that
g(f(x)) =w.Ify:=f(x), theny∈Bandg(y)=w.Thusgis surjective.
23. We havex∈f
−1
(g
−1
(H))⇐⇒f(x)∈g
−1
(H)⇐⇒g(f(x))∈H⇐⇒x∈
(g◦f)
−1
(H).
24. Ifg(f(x)) =xfor allx∈D(f), theng◦fis injective, and Exercise 22(a)
implies thatfis injective onD(f). Iff(g(y)) =yfor ally∈D(g), then
Exercise 22(b) implies thatfmapsD(f)ontoD(g).Thusfis a bijection of
D(f)ontoD(g), andg=f
−1
.
Section 1.2
The method of proof known as Mathematical Induction is used frequently in real
analysis, but in many situations the details follow a routine patterns and are

4 Bartle and Sherbert
left to the reader by means of a phrase such as: “The proof is by Mathematical
Induction”. Since may students have only a hazy idea of what is involved, it may
be a good idea to spend some time explaining and illustrating what constitutes a
proof by induction.
Pains should be taken to emphasize that the induction hypothesis doesnot
entail “assuming what is to be proved”. The inductive step concerns the validity
of going from the assertion fork∈Nto that fork+ 1. The truth of falsity of the
individual assertion is not an issue here.
Sample Assignment: Exercises 1, 2, 6, 11, 13, 14, 20.
Partial Solutions:
1. The assertion is true forn= 1 because 1/(1·2)=1/(1+1).If it is true
forn=k, then it follows fork+ 1 becausek/(k+1)+1/[(k+ 1)(k+ 2)] =
(k+1)/(k+ 2).
2. The statement is true forn= 1 because [
1
2
·1·2]
2
=1=1
3
. For the inductive
step, use the fact that
Φ
1
2
k(k+1)
Ψ
2
+(k+1)
3
=
Φ
1
2
(k+ 1)(k+2)
Ψ
2
.
3. It is true forn= 1 since 3 = 4−1.If the equality holds forn=k, then
add 8(k+1)−5=8k+ 3 to both sides and show that (4k
2
−k)+(8k+3)=
4(k+1)
2
−(k+ 1) to deduce equality for the casen=k+1.
4. It is true forn= 1 since 1 = (4−1)/3. If it is true forn=k, then add
(2k+1)
2
to both sides and use some algebra to show that
1
3
(4k
3
−k)+(2k+1)
2
=
1
3
[4k
3
+12k
2
+11k+3]=
1
3
[4(k+1)
3
−(k+ 1)],
which establishes the casen=k+1.
5. Equality holds forn= 1 since 1
2
=(−1)
2
(1·2)/2. The proof is completed by
showing (−1)
k+1
[k(k+ 1)]/2+(−1)
k+2
(k+1)
2
=(−1)
k+2
[(k+ 1)(k+ 2)]/2.
6. Ifn= 1, then 1
3
+5·1 = 6 is divisible by 6. Ifk
3
+5kis divisible by 6,
then (k+1)
3
+5(k+1)=(k
3
+5k)+3k(k+ 1) + 6 is also, becausek(k+1)
is always even (why?) so that 3k(k+ 1) is divisible by 6, and hence the sum
is divisible by 6.
7. If 5
2k
−1 is divisible by 8, then it follows that 5
2(k+1)
−1=(5
2k
−1)+24·5
2k
is also divisible by 8.
8. 5
k+1
−4(k+1)−1=5·5
k
−4k−5=(5
k
−4k−1) + 4(5
k
−1). Now show that
5
k
−1 is always divisible by 4.
9. Ifk
3
+(k+1)
3
+(k+2)
3
is divisible by 9, then (k+1)
3
+(k+2)
3
+(k+3)
3
=
k
3
+(k+1)
3
+(k+2)
3
+9(k
2
+3k+ 3) is also divisible by 9.
10. The sum is equal ton/(2n+1).

Chapter 1 — Preliminaries 5
11. The sum is 1+3+···+(2n−1) =n
2
. Note thatk
2
+(2k+1)=(k+1)
2
.
12. Ifn
0>1, letS 1:={n∈N:n−n 0+1∈S}Apply 1.2.2 to the setS 1.
13. Ifk<2
k
, thenk+1<2
k
+1<2
k
+2
k
= 2(2
k
)=2
k+1
.
14. Ifn= 4, then 2
4
=16<24=4!.If 2
k
<k! and ifk≥4,then 2
k+1
=2·2
k
<
2·k!<(k+1)·k!=(k+ 1)!. [Note that the inductive step is valid when-
ever 2<k+ 1, includingk=2,3,even though the statement is false for these
values.]
15. Forn= 5 we have 7≤2
3
.Ifk≥5 and 2k−3≤2
k−2
, then 2(k+1)−3=
(2k−3)+2≤2
k−2
+2
k−2
=2
(k+1)−2
.
16. It is true forn= 1 andn≥5, but false forn=2,3,4. The inequality
2k+1<2
k
, wich holds fork≥3,is needed in the induction argument. [The
inductive step is valid forn=3,4 even though the inequalityn
2
<2
n
is false
for these values.]
17.m= 6 trivially dividesn
3
−nforn= 1, and it is the largest integer to divide
2
3
−2=6. Ifk
3
−kis divisible by 6, then sincek
2
+kis even (why?), it
follows that (k+1)
3
−(k+1)=(k
3
−k)+3(k
2
+k) is also divisible by 6.
18.
√
k+1/
√
k+1=(
√
k
√
k+1+1)/
√
k+1>(k+1)/
√
k+1=
√
k+1.
19. First note that since 2∈S, then the number 1 = 2−1 belongs toS.Ifm/∈S,
thenm<2
m
∈S,so2
m
−1∈S, etc.
20. If 1≤x
k−1≤2 and 1≤x k≤2, then 2≤x k−1+xk≤4, so that 1≤x k+1=
(x
k−1+xk)/2≤2.
Section 1.3
Every student of advanced mathematics needs to know the meaning of the words
“ﬁnite”, “inﬁnite”, “countable” and “uncountable”. For most students at this
level it is quite enough to learn the deﬁnitions and read the statements of the
theorems in this section, but to skip the proofs. Probably every instructor will
want to show thatQis countable andRis uncountable (see Section 2.5).
Some students will not be able to comprehend that proofs are necessary for
“obvious” statements about ﬁnite sets. Others will ﬁnd the material absolutely
fascinating and want to prolong the discussion forever. The teacher must avoid
getting bogged down in a protracted discussion of cardinal numbers.
Sample Assignment: Exercises 1, 5, 7, 9, 11.
Partial Solutions:
1. IfT
1 =∅is ﬁnite, then the deﬁnition of a ﬁnite set applies toT 2=Nnfor
somen.Iffis a bijection ofT
1ontoT 2, and ifgis a bijection ofT 2ontoN n,
then (by Exercise 1.1.21) the compositeg◦fis a bijection ofT
1ontoN n,so
thatT
1is ﬁnite.

6 Bartle and Sherbert
2. Part (b) Letfbe a bijection ofN
montoAand letC={f(k)}for some
k∈N
m. DeﬁnegonN m−1byg(i):=f(i) fori=1,...,k−1, andg(i):=
f(i+ 1) fori=k,...,m−1. Thengis a bijection ofN
m−1ontoA\C. (Why?)
Part (c) First note that the union of two ﬁnite sets is a ﬁnite set. Now note
that ifC/Bwere ﬁnite, thenC=B∪(C\B) would also be ﬁnite.
3. (a) The element 1 can be mapped into any of the three elements ofT, and
2 can then be mapped into any of the two remaining elements ofT, after
which the element 3 can be mapped into only one element ofT.Hence there
are 6 = 3·2·1 diﬀerent injections ofSintoT.
(b) Supposeamaps into 1. Ifbalso maps into 1, thencmust map into 2; ifb
maps into 2, thenccan map into either 1 or 2. Thus there are 3 surjections
that mapainto 1, and there are 3 other surjections that mapainto 2.
4.f(n):=2n+13,n∈N.
5.f(1) := 0,f(2n):=n, f(2n+1):=−nforn∈N.
6. The bijection of Example 1.3.7(a) is one example. Another is the shift deﬁned
byf(n):=n+ 1 that mapsNontoN\{1}.
7. IfT
1is denumerable, takeT 2=N.Iffis a bijection ofT 1ontoT 2, and ifg
is a bijection ofT
2ontoN, then (by Exercise 1.1.21)g◦fis a bijection ofT 1
ontoN, so thatT 1is denumerable.
8. LetA
n:={n}forn∈N,so
ﬀ
A n=N.
9. IfS∩T=∅andf:N→S, g:N→Tare bijections ontoSandT, respectively,
leth(n):=f((n+1)/2) ifnis odd andh(n):=g(n/2) ifnis even. It is readily
seen thathis a bijection ofNontoS∪T; henceS∪Tis denumerable. What
ifS∩T =∅?
10. (a)m+n−1=9 andm= 6 implyn= 4. Thenh(6,4) =
1
2
·8·9+6=42.
(b)h(m,3) =
1
2
(m+ 1)(m+2)+m= 19, so thatm
2
+5m−36=0. Thus
m=4.
11. (a)P({1,2})={∅,{1},{2},{1,2}}has 2
2
= 4 elements.
(b)P({1,2,3}) has 2
3
= 8 elements.
(c)P({1,2,3,4}) has 2
4
= 16 elements.
12. LetS
n+1:={x 1,...,xn,xn+1}=S n∪{x n+1}haven+ 1 elements. Then a
subset ofS
n+1either (i) containsx n+1, or (ii) does not containx n+1. The
induction hypothesis implies that there are 2
n
subsets of type (i), since each
such subset is the union of{x
n+1}and a subset ofS n. There are also 2
n
subsets of type (ii). Thus there is a total of 2
n
+2
n
=2·2
n
=2
n+1
subsets
ofS
n+1.
13. For eachm∈N, the collection of all subsets ofN
mis ﬁnite. (See Exercise 12.)
Every ﬁnite subset ofNis a subset ofN
mfor a suﬃciently largem. Therefore
Theorem 1.3.12 implies thatF(N)=
ﬀ
∞
m=1
P(Nm) is countable.

CHAPTER 2
THE REAL NUMBERS
Students will be familiar with much of the factual content of the ﬁrst few sections,
but the process of deducing these facts from a basic list of axioms will be new
to most of them. The ability to construct proofs usually improves gradually
during the course, and there are much more signiﬁcant topics forthcoming. A few
selected theorems should be proved in detail, since some experience in writing
formal proofs is important to students at this stage. However, one should not
spend too much time on this material.
Sections 2.3 and 2.4 on the Completeness Property form the heart of this
chapter.These sections should be covered thoroughly.Also the Nested Intervals
Property in Section 2.5 should be treated carefully.
Section 2.1
One goal of Section 2.1 is to acquaint students with the idea of deducing conse-
quences from a list of basic axioms. Students who have not encountered this type
of formal reasoning may be somewhat uncomfortable at ﬁrst, since they often
regard these results as “obvious”. Since there is much more to come, a sampling
of results will suﬃce at this stage, making it clear that it is only a sampling.
The classic proof of the irrationality of
√
2 should certainly be included in the
discussion, and students should be asked to modify this argument for
√
3,etc.
Sample Assignment: Exercises 1(a,b), 2(a,b), 3(a,b), 6, 13, 16(a,b), 20, 23.
Partial Solutions:
1. (a) Apply appropriate algebraic properties to getb=0+b=(−a+a)+b=
−a+(a+b)=−a+0=−a.
(b) Apply (a) to (−a)+a= 0 withb=ato conclude thata=−(−a).
(c) Apply (a) to the equationa+(−1)a=a(1+(−1)) =a·0 = 0 to conclude
that (−1)a=−a.
(d) Apply (c) witha=−1 to get (−1)(−1) =−(−1). Then apply (b) with
a= 1 to get (−1)(−1)=1.
2. (a)−(a+b)=(−1)(a+b)=(−1)a+(−1)b=(−a)+(−b).
(b) (−a)·(−b)=((−1)a)·((−1)b)=(−1)(−1)(ab)=ab.
(c) Note that (−a)(−(1/a)) =a(1/a)=1.
(d)−(a/b)=(−1)(a(1/b))=((−1)a)(1/b)=(−a)/b.
3. (a) Add−5 to both sides of 2x+5=8 and use(A2),(A4),(A3) to get 2x=3.
Then multiply both sides by 1/2 to getx=3/2.
(b) Writex
2
−2x=x(x−2) = 0 and apply Theorem 2.1.3(b). Alternatively,
note thatx= 0 satisﬁes the equation, and ifx = 0, then multiplication by
1/xgivesx=2.
7

8 Bartle and Sherbert
(c) Add−3 to both sides and factor to getx
2
−4=(x−2)(x+2)=0. Now
apply 2.1.3(b) to getx=2 orx=−2.
(d) Apply 2.1.3(b) to show that (x−1)(x+ 2) = 0 if and only ifx=1 or
x=−2.
4. Clearlya= 0 satisﬁesa·a=a.Ifa = 0 anda·a=a, then (a·a)(1/a)=a(1/a),
so thata=a(a(1/a)) =a(1/a)=1.
5. If (1/a)(1/b) is multiplied byab,the result is 1. Therefore, Theorem 2.1.3(a)
implies that 1/(ab)=(1/a)(1/b).
6. Note that ifq∈Zand if 3q
2
is even, thenq
2
is even, so thatqis even. Hence,
if (p/q)
2
= 6, then it follows thatpis even, sayp=2m, whence 2m
2
=3q
2
,so
thatqis also even.
7. Ifp∈N, there are three possibilities: for somem∈N∪{0}, (i)p=3m,
(ii)p=3m+ 1, or (iii)p=3m+ 2. In either case (ii) or (iii), we havep
2
=
3h+ 1 for someh∈N∪{0}.
8. (a) Letx=m/n, y=p/q, wherem, n =0,p,q = 0 are integers. Thenx+y=
(mq+np)/nqandxy=mp/nqare rational.
(b) Ifs:=x+y∈Q, theny=s−x∈Q, a contradiction. Ift:=xy∈Qand
x = 0, theny=t/x∈Q, a contradiction.
9. (a) Ifx
1=s1+t1
√
2 andx 2=s2+t2
√
2 are inK, thenx 1+x2=
(s
1+s2)+(t 1+t2)
√
2 andx 1x2=(s1s2+2t 1t2)+(s 1t2+s2t1)
√
2 are also
inK.
(b) Ifx=s+t
√
2 = 0 is inK, thens−t
√
2 = 0 (why?) and
1
x
=
s−t
√
2
(s+t
√
2)(s−t
√
2)
=
π
s
s
2
−2t
2
α
−
π
t
s
2
−2t
2
α
√
2
is inK. (Use Theorem 2.1.4.)
10 (a) Ifc=d, then 2.1.7(b) impliesa+c<b+d.Ifc<d, thena+c<
b+c<b+d.
(b) Ifc=d= 0, thenac=bd=0. Ifc>0, then 0<acby the Trichotomy
Property andac<bcfollows from 2.1.7(c). If alsoc≤d, thenac≤ad < bd.
Thus 0≤ac≤bdholds in all cases.
11. (a) Ifa>0, thena = 0 by the Trichotomy Property, so that 1/aexists. If
1/a= 0, then 1 =a·(1/a)=a·0 = 0, which contradicts (M3). If 1/a <0, then
2.1.7(c) implies that 1 =a(1/a)<0, which contradicts 2.1.8(b). Thus 1/a >0,
and 2.1.3(a) implies that 1/(1/a)=a.
(b) Ifa<b, then 2a=a+a<a+b, and alsoa+b<b+b=2b. Therefore,
2a<a+b<2b, which, since
1
2
>0 (by 2.1.8(c) and part (a)), implies that
a<
1
2
(a+b)<b.
12. Leta= 1 andb=2. Ifc=−3 andd=−1, thenac<bd. On the other hand,
ifc=−3 andd=−2,thenbd<ac. (Many other examples are possible.)

Chapter 2 — The Real Numbers 9
13. Ifa = 0, then 2.1.8(a) implies thata
2
>0; sinceb
2
≥0, it follows that
a
2
+b
2
>0.
14. If 0≤a<b, then 2.1.7(c) impliesab<b
2
.Ifa= 0, then 0 =a
2
=ab<b
2
.
Ifa>0, thena
2
<abby 2.1.7(c). Thusa
2
≤ab<b
2
.Ifa=0,b= 1, then
0=a
2
=ab<b=1.
15. (a) If 0<a<b, then 2.1.7(c) implies that 0<a
2
<ab<b
2
. Then by Example
2.1.13(a), we infer thata=
√
a
2
<
√
ab <
√
b
2
=b.
(b) If 0<a<b, thenab >0 so that 1/ab >0, and thus 1/a−1/b=
(1/ab)(b−a)>0.
16. (a) To solve (x−4)(x+1)>0, look at two cases. Case 1:x−4>0 and
x+1>0, which givesx>4. Case 2:x−4<0 andx+1<0, which gives
x<−1. Thus we have{x:x>4orx<−1}.
(b) 1<x
2
<4 has the solution set{x:1<x<2or−2<x<−1}.
(c) The inequality is 1/x−x=(1−x)(1 +x)/x <0. Ifx>0, this is equiva-
lentto(1−x)(1 +x)<0, which is satisﬁed ifx>1. Ifx<0, then we solve
(1−x)(1 +x)>0, and get−1<x<0. Thus we get{x:−1<x<0orx>1}
(d) the solution set is{x:x<0orx>1}.
17. Ifa>0, we can takeε
0:=a>0 and obtain 0<ε 0≤a, a contradiction.
18. Ifb<aand ifε
0:= (a−b)/2, thenε 0>0 anda=b+2ε 0>b+ε 0.
19. The inequality is equivalent to 0≤a
2
−2ab+b
2
=(a−b)
2
.
20. (a) If 0<c<1, then 2.1.7(c) implies that 0<c
2
<c, whence 0<c
2
<c<1.
(b) Sincec>0, then 2.1.7(c) implies thatc<c
2
, whence 1<c<c
2
.
21. (a) LetS:={n∈N:0<n<1}.IfSis not empty, the Well-Ordering Property
ofNimplies there is a least elementminS. However, 0<m<1 implies that
0<m
2
<m, and sincem
2
is also inS, this is a contradiction to the fact that
mis the least element ofS.
(b) Ifn=2p=2q−1 for somep, qinN, then 2(q−p) = 1, so that 0<q−p<1.
This contradicts (a).
22. (a) Letx:=c−1>0 and apply Bernoulli’s Inequality 2.1.13(c) to getc
n
=
(1 +x)
n
≥1+nx≥1+x=cfor alln∈N, andc
n
>1+x=cforn>1.
(b) Letb:= 1/cand use part (a).
23. If 0<a<banda
k
<b
k
, then 2.1.7(c) implies thata
k+1
<ab
k
<b
k+1
so
Induction applies. Ifa
m
<b
m
for somem∈N, the hypothesis that 0<b≤a
leads to a contradiction.
24. (a) Ifm>n, thenk:=m−n∈N, so Exercise 22(a) implies thatc
k
≥c>1.
But sincec
k
=c
m−n
, this implies thatc
m
>c
n
. Conversely, the hypothesis
thatc
m
>c
n
andm≤nlead to a contradiction.
(b) Letb:= 1/cand use part (a).

10 Bartle and Sherbert
25. Letb:=c
1/mn
. We claim thatb>1; for ifb≤1, then Exercise 22(b) implies
that 1<c=b
mn
≤b≤1, a contradiction. Therefore Exercise 24(a) implies
thatc
1/n
=b
m
>b
n
=c
1/m
if and only ifm>n.
26. Fixm∈Nand use Mathematical Induction to prove thata
m+n
=a
m
a
n
and
(a
m
)
n
=a
mn
for alln∈N. Then, for a givenn∈N, prove that the equalities
are valid for allm∈N.
Section 2.2
The notion of absolute value of a real number is deﬁned in terms of the basic order
properties ofR. We have put it in a separate section to give it emphasis. Many
students need extra work to become comfortable with manipulations involving
absolute values, especially when inequalities are involved.
We have also used this section to give students an early introduction to the
notion of theε-neighborhood of a point. As a preview of the role of
ε-neighborhoods, we have recast Theorem 2.1.9 in terms ofε-neighborhhoods in
Theorem 2.2.8.
Sample Assignment: Exercises 1, 4, 5, 6(a,b), 8(a,b), 9, 12(a,b), 15.
Partial Solutions:
1. (a) Ifa≥0, then|a|=a=
√
a
2
;ifa<0, then|a|=−a=
√
a
2
.
(b) It suﬃces to show that|1/b|=1/|b|forb = 0 (why?). Ifb>0, then
1/b >0 (why?), so that|1/b|=1/b=1/|b|.Ifb<0, then 1/b <0, so that
|1/b|=−(1/b)=1/(−b)=1/|b|.
2. First show thatab≥0 if an only if|ab|=ab. Then show that (|a|+|b|)
2
=
(a+b)
2
if and only if|ab|=ab.
3. Ifx≤y≤z, then|x−y|+|y−z|=(y−x)+(z−y)=z−x=|z−x|.To
establish the converse, show thaty<xandy>zare impossible. For example,
ify<x≤z, it follows from what we have shown and the given relationship
that|x−y|= 0, so thaty=x, a contradiction.
4.|x−a|<ε⇐⇒ −ε<x−a<ε⇐⇒a−ε<x<a+ε.
5. Ifa<x<band−b<−y<−a, it follows thata−b<x−y<b−a. Since
a−b=−(b−a),the argument in 2.2.2(c) gives the conclusion|x−y|<b−a.
The distance betweenxandyis less than or equal tob−a.
6. (a)|4x−5|≤13⇐⇒ −13≤4x−5≤13⇐⇒ −8≤4x≤18⇐⇒ −2≤
x≤9/2.
(b)|x
2
−1|≤3⇐⇒ −3≤x
2
−1≤3⇐⇒ −2≤x
2
≤4⇐⇒0≤x
2
≤4⇐⇒
−2≤x≤2.
7. Case 1:x≥2⇒(x+1)+(x−2)=2x−1=7, sox=4.
Case 2:−1<x<2⇒(x+1)+(2−x)=3 =7,so no solution.
Case 3:x≤−1⇒(−x−1)+(2−x)=−2x+1=7, sox=−3.
Combining these cases, we getx=4 orx=−3.

Chapter 2 — The Real Numbers 11
8. (a) Ifx>1/2, thenx+1=2x−1, so thatx=2. Ifx≤1/2, thenx+1=
−2x+ 1, so thatx= 0. There are two solutions{0,2}.
(b) Ifx≥5, the equation impliesx=−4, so no solutions. Ifx<5, thenx=2.
9. (a) Ifx≥2, the inequality becomes−2≤1. Ifx≤2, the inequality isx≥1/2,
so this case contributes 1/2≤x≤2. Combining the cases gives us allx≥1/2.
(b)x≥0 yieldsx≤1/2, so that we get 0≤x≤1/2.x≤0 yieldsx≥−1, so
that−1≤x≤0. Combining cases, we get−1≤x≤1/2.
10. (a) Either consider the three cases:x<−1,−1≤x≤1 and 1<x; or, square
both sides to get−2x>2x. Either approach givesx<0.
(b) Consider the three casesx≥0,−1≤x<0 andx<−1toget−3/2<
x<1/2.
11.y=f(x) wheref(x):=−1 forx<0,f(x):=2x−1 for 0≤x≤1, andf(x):=1
forx>1.
12. Case 1:x≥1⇒4<(x+2)+(x−1)<5,so 3/2<x<2.
Case 2:−2<x<1⇒4<(x+2)+(1−x)<5, so there is no solution.
Case 3:x<−2⇒4<(−x−2)+(1−x)<5, so−3<x<−5/2.
Thus the solution set is{x:−3<x<−5/2or3/2<x<2}.
13.|2x−3|<5⇐⇒ −1<x<4, and|x+1|>2⇐⇒x<−3orx>1. The two
inequalities are satisﬁed simultaneously by points in the intersection{x:
1<x<4}.
14. (a)|x|=|y|⇐⇒x
2
=y
2
⇐⇒(x−y)(x+y)=0⇐⇒y=xory=−x.Thus
{(x, y):y=xory=−x}.
(b) Consider four cases. Ifx≥0,y≥0, we get the line segment joining the
points (0, 1) and (1, 0). Ifx≤0,y≥0, we get the line segment joining (−1,0)
and (0,1), and so on.
(c) The hyperbolasy=2/xandy=−2/x.
(d) Consider four cases corresponding to the four quadrants. The graph
consists of a portion of a line segment in each quadrant. For example, if
x≥0,y≥0, we obtain the portion of the liney=x−2 in this quadrant.
15. (a) Ify≥0, then−y≤x≤yand we get the region in the upper half-plane on
or between the linesy=xandy=−x.Ify≤0, then we get the region in the
lower half-plane on or between the linesy=xandy=−x.
(b) This is the region on and inside the diamond with vertices (1, 0), (0, 1),
(−1,0) and (0,−1).
16. For the intersection, letγbe the smaller ofεandδ. For the union, letγbe
the larger ofεandδ.
17. Choose anyε>0 such thatε<|a−b|.
18. (a) Ifa≤b, then max{a, b}=b=
1
2
[a+b+(b−a)] and min{a, b}=a=
1
2
[a+b−(b−a)].
(b) Ifa= min{a, b, c}, then min{min{a, b},c}=a= min{a, b, c}. Similarly, if
borcis min{a, b, c}.

12 Bartle and Sherbert
19. Ifa≤b≤c, then mid{a, b, c}=b= min{b, c, c}= min{max{a, b}, max{b, c},
max{c, a}}. The other cases are similar.
Section 2.3
This section completes the description of the real number system by introducing
the fundamental completeness property in the form of the Supremum Property.
This property is vital to real analysis and students should attain a working under-
standing of it. Eﬀort expended in this section and the one following will be richly
rewarded later.
Sample Assignment: Exercises 1, 2, 5, 6, 9, 10, 12, 14.
Partial Solutions:
1. Any negative number or 0 is a lower bound. For anyx≥0, the larger number
x+ 1 is inS
1, so thatxis not an upper bound ofS 1. Since 0≤xfor allx∈S 1,
thenu= 0 is a lower bound ofS
1.Ifv>0, thenvis not a lower bound ofS 1
becausev/2∈S 1andv/2<v. Therefore infS 1=0.
2.S
2has lower bounds, so that infS 2exists. The argument used forS 1also
shows that infS
2= 0, but that infS 2does not belong toS 2.S2does not
have upper bounds, so that supS
2does not exists.
3. Since 1/n≤1 for alln∈N, then 1 is an upper bound forS
3. But1isa
member ofS
3, so that 1 = supS 3. (See Exercise 7 below.)
4. supS
4= 2 and infS 4=1/2. (Note that both are members ofS 4.)
5. It is interesting to compare algebraic and geometric approaches to these
problems.
(a) infA=−5/2, supAdoes not exist,
(b) supB= 2, infB=−1,
(c) supC= 1, infBdoes not exist,
(d) supD=1+
√
6, infD=1−
√
6.
6. IfSis bounded below, thenS
ﬃ
:={−s:s∈S}is bounded above, so that
u:= supS
ﬃ
exists. Ifv≤sfor alls∈S, then−v≥−sfor alls∈S, so that
−v≥u, and hencev≤−u. Thus infS=−u.
7. Letu∈Sbe an upper bound ofS.Ifvis another upper bound ofS, then
u≤v. Henceu= supS.
8. Ift>uandt∈S, thenuis not an upper bound ofS.
9. Letu:= supS. Sinceuis an upper bound ofS,so isu+1/nfor alln∈N.
Sinceuis the supremum ofSandu−1/n<u, then there existss
0∈Swith
u−1/n<s
0, whenceu−1/nis not an upper bound ofS.
10. Letu:= supA,v:= supBandw:= sup{u, v}. Thenwis an upper bound of
A∪B, because ifx∈A, thenx≤u≤w, and ifx∈B, thenx≤v≤w.Ifzis

Chapter 2 — The Real Numbers 13
any upper bound ofA∪B, thenzis an upper bound ofAand ofB, so that
u≤zandv≤z. Hencew≤z. Therefore,w= sup(A∪B).
11. Since supSis an upper bound ofS, it is an upper bound ofS
0, and hence
supS
0≤supS.
12. Consider two cases. Ifu≥s
∗
, thenu= sup(S∪{u}).Ifu<s
∗
, then there
existss∈Ssuch thatu<s≤s
∗
, so thats
∗
= sup(S∪{u}).
13. IfS
1:={x 1},show thatx 1= supS 1.IfS k:={x 1,...,xk}is such that sup
S
k∈Sk, then preceding exercise implies that sup{x 1,...,xk,xk+1}is the
larger of supS
kandx k+1and so is inS k+1.
14. Ifw= infSandε>0, thenw+εis not a lower bound so that there existst
inSsuch thatt<w+ε.Ifwis a lower bound ofSthat satisﬁes the stated
condition, and ifz>w, letε=z−w>0. Then there istinSsuch that
t<w+ε=z, so thatzis not a lower bound ofS.Thus,w= infS.
Section 2.4
This section exhibits how the supremum is used in practice, and contains some
important properties ofRthat will often be used later. The Archimedean Proper-
ties 2.4.3–2.4.6 and the Density Properties 2.4.8 and 2.4.9 are the most signiﬁcant.
The exercises also contain some results that will be used later.
Sample Assignment: Exercises 1, 2, 4(b), 5, 7, 10, 12, 13, 14.
Partial Solutions:
1. Since 1−1/n <1 for alln∈N, the number 1 is an upper bound. To show
that 1 is the supremum, it must be shown that for eachε>0 there exists
n∈Nsuch that 1−1/n >1−ε, which is equivalent to 1/n<ε. Apply the
Archimedean Property 2.4.3 or 2.4.5.
2. infS=−1 and supS= 1. To see the latter note that 1/n−1/m≤1 for all
m, n∈N. On the other hand ifε>0 there existsm∈Nsuch that 1/m<ε,
so that 1/1−1/m >1−ε.
3. Suppose thatu∈Ris not the supremum ofS. Then either (i)uis not an
upper bound ofS(so that there existss
1∈Swithu<s 1, whence we take
n∈Nwith 1/n<s
1−uto show thatu+1/nis not an upper bound ofS), or
(ii) there exists an upper boundu
1ofSwithu 1<u(in which case we take
1/n<u−u
1to show thatu−1/nis not an upper bound ofS).
4. (a) Letu:= supSanda>0. Thenx≤ufor allx∈S, whenceax≤aufor all
x∈S, whence it follows thatauis an upper bound ofaS.Ifvis another upper
bound ofaS, thenax≤vfor allx∈S, whencex≤v/afor allx∈S, showing
thatv/ais an upper bound forSso thatu≤v/a, from which we conclude
thatau≤v. Thereforeau= sup(aS). The statement about the inﬁmum is
proved similarly.

14 Bartle and Sherbert
(b) Letu:= supSandb<0. Ifx∈S, then (sinceb<0)bu≤bxso that
buis a lower bound ofbS.Ifv≤bxfor allx∈S, thenx≤v/b(sinceb<0),
so thatv/bis an upper bound forS. Henceu≤v/bwhencev≤bu. Therefore
bu= inf(bS).
5. Ifx∈S, then 0≤x≤u, so thatx
2
≤u
2
which implies supT≤u
2
.Iftis any
upper bound ofT, thenx∈Simpliesx
2
≤tso thatx≤
√
t. It follows that
u≤
√
t, so thatu
2
≤t.Thusu
2
≤supT.
6. Letu:= supf(X). Thenf(x)≤ufor allx∈X, so thata+f(x)≤a+ufor
allx∈X, whence sup{a+f(x):x∈X}≤a+u.Ifw<a+u, thenw−a<u,
so that there existsx
w∈Xwithw−a<f(x w), whencew<a+f(x w), and
thuswis not an upper bound for{a+f(x):x∈X}.
7. Letu:= supS,v:= supB,w:= sup(A+B). Ifx∈Aandy∈B, then
x+y≤u+v, so thatw≤u+v.Now,ﬁx y∈Band note thatx≤w−y
for allx∈A;thusw−yis an upper bound forAso thatu≤w−y. Then
y≤w−ufor ally∈B,sov≤w−uand henceu+v≤w. Combining these
inequalities, we havew=u+v.
8. Ifu:= supf(X) andv:= supg(X), thenf(x)≤uandg(x)≤vfor allx∈X,
whencef(x)+g(x)≤u+vfor allx∈X.Thusu+vis an upper bound
for the set{f(x)+g(x):x∈X}. Therefore sup{f(x)+g(x):x∈X}≤
u+v.
9. (a)f(x)=2x+1,inf{f(x):x∈X}=1.
(b)g(y)=y, sup{g(y):y∈Y}=1.
10. (a)f(x)=1 forx∈X. (b)g(y)=0 fory∈Y.
11. Ifx∈X,y∈Y, theng(y)≤h(x, y)≤f(x). If we ﬁxy∈Yand take the
inﬁmum overx∈X, then we getg(y)≤inf{f(x):x∈X}for eachy∈Y.
Now take the supremum overy∈Y.
12. LetS:={h(x, y):x∈X, y∈Y}. We haveh(x, y)≤F(x) for allx∈X, y∈Y
so that supS≤sup{F(x):x∈X}.Ifw<sup{F(x):x∈X}, then there
existsx
0∈Xwithw<F(x 0) = sup{h(x 0,y):y∈Y}, whence there exists
y
0∈Ywithw<h(x 0,y0).Thuswis not an upper bound ofS, and so
w<supS.Since this is true for anywsuch thatw<sup{F(x):x∈X},
we conclude that sup{F(x):x∈X}≤supS.
13. Ifx∈Z, taken:=x+1. Ifx/∈Z, we have two cases: (i)x>0 (which is
covered by Cor. 2.4.6), and (ii)x<0. In case (ii), letz:=−xand use 2.4.6.
Ifn
1<n2are integers, thenn 1≤n2−1 so the sets{y:n 1−1≤y<n 1}and
{y:n
2−1≤y<n 2}are disjoint; thus the integernsuch thatn−1≤x<n
is unique.
14. Note thatn<2
n
(whence 1/2
n
<1/n) for anyn∈N.
15. LetS
3:={s∈R:0≤s, s
2
<3}. Show thatS 3is nonempty and bounded
by 3 and lety:= supS
3.Ify
2
<3 and 1/n <(3−y
2
)/(2y+ 1) show that

Chapter 2 — The Real Numbers 15
y+1/n∈S
3.Ify
2
>3 and 1/m <(y
2
−3)/2yshow thaty−1/m∈S 3.
Thereforey
2
=3.
16. Case 1: Ifa>1, letS
a:={s∈R:0≤s, s
2
<a}. Show thatS ais nonempty
and bounded above byaand letz:= supS
a. Now show thatz
2
=a.
Case 2: If 0<a<1, there existsk∈Nsuch thatk
2
a>1 (why?). Ifz
2
=k
2
a,
then (z/k)
2
=a.
17. ConsiderT:={t∈R:0≤t, t
3
<2}.Ift>2, thent
3
>2 so thatt/∈T. Hence
y:= supTexists. Ify
3
<2, choose 1/n <(2−y
3
)/(3y
2
+3y+ 1) and show
that (y+1/n)
3
<2, a contradiction, and so on.
18. Ifx<0<y, then we can taker=0. Ifx<y<0, we apply 2.4.8 to obtain a
rational number between−yand−x.
19. There existsr∈Qsuch thatx/u<r<y/u.
Section 2.5
Another important consequence of the Supremum Property ofRis the Nested
Intervals Property 2.5.2. It is an interesting fact that if we assume the validity of
boththe Archimedean Property 2.4.3 and the Nested Intervals Property, then we
can prove the Supremum Property. Hence these two properties could be taken
as the completeness axiom forR. However, establishing this logical equivalence
would consume valuable time and not signiﬁcantly advance the study of real anal-
ysis, so we will not do so. (There are other properties that could be taken as the
completeness axiom.)
The discussion of binary and decimal representations is included to give the
student a concrete illustration of the rather abstract ideas developed to this point.
However, this material is not vital for what follows and can be omitted or treated
lightly. We have kept this discussion informal to avoid getting buried in technical
details that are not central to the course.
Sample Assignment: Exercises 3, 4, 5, 6, 7, 8, 10, 11.
Partial Solutions:
1. Note that [a, b]⊆[a
ﬃ
,b
ﬃ
] if and only ifa
ﬃ
≤a≤b≤b
ﬃ
.
2.Shas an upper boundband a lower boundaif and only ifSis contained in
the interval [a, b].
3. Since infSis a lower bound forSand supSis an upper bound forS, then
S⊆I
S. Moreover, ifS⊆[a, b],thenais a lower bound forSandbis an
upper bound forS, so that [a, b]⊇I
S.
4. Becausezis neither a lower bound or an upper bound ofS.
5. Ifz∈R, thenzis not a lower bound ofSso there existsx
z∈Ssuch that
x
z≤z. Alsozis not an upper bound ofSso there existsy z∈Ssuch that
z≤y
z. Sincezbelongs to [x z,yz],it follows from the property (1) thatz∈S.

16 Bartle and Sherbert
But sincez∈Ris arbitrary, we conclude thatR⊆S, whence it follows that
S=R=(−∞,∞).
6. Since [a
n,bn]=In⊇In+1=[an+1,bn+1], it follows as in Exercise 1 that
a
n≤an+1≤bn+1≤bn. Therefore we havea 1≤a2≤ ··· ≤a n≤ ···and
b
1≥b2≥ ··· ≥b n≥ ···.
7. Since 0∈I
nfor alln∈N, it follows that 0∈
ﬃ
∞
n=1
In. On the other hand
ifu>0, then Corollary 2.4.5 implies that there existsn∈Nwith 1/n<u,
whenceu/∈[0,1/n]=I
n. Therefore, such audoes not belong to this
intersection.
8. Ifx>0, then there existsn∈Nwith 1/n<x, so thatx/∈J
n.Ify≤0, then
y/∈J
1.
9. Ifz≤0, thenz/∈K
1.Ifw>0, then it follows from the Archimedean Property
that there existsn
w∈Nwithw/∈(n w,∞)=K nw.
10. Letη:= inf{b
n:n∈N}; we claim thata n≤ηfor alln. Fixn∈N; we will
show thata
nis a lower bound for the set{b k:k∈N}. We consider two cases.
(j) Ifn≤k, then sinceI
n⊇Ik, we havea n≤ak≤bk. (jj) Ifk<n, then since
I
k⊇In, we havea n≤bn≤bk. Thereforea n≤bkfor allk∈N, so thata nis
a lower bound for{b
k:k∈N}and soa n≤η. In particular, this shows that
η∈[a
n,bn] for alln, so thatη∈
ﬃ
I n.
In view of 2.5.2, we have [ξ,η]⊂I
nfor alln, so that [ξ,η]⊆
ﬃ
I n. Con-
versely, ifz∈I
nfor alln, thena n≤z≤b nfor alln, whence it follows that
ξ= sup{a
n}≤z≤inf{b n}=η. Therefore
ﬃ
I n⊆[ξ,η] and so equality holds.
11. Ifn∈N, letc
n:=a1/2+a 2/2
2
+···+a n/2
n
andd n:=a1/2+a 2/2
2
+···+
(a
n+1)/2
n
, and letJ n:= [cn,dn]. Sincec n≤cn+1≤dn+1≤dnforn∈N, the
intervalsJ
nform a nested sequence.
12.
3
8
=(.011000···) 2=(.010111···) 2.
7
16
=(.0111000···) 2=(.0110111···) 2.
13. (a)
1
3
≈(.0101) 2 (b)
1
3
=(.010101···) 2, the block 01 repeats.
14. We may assume thata
n =0. Ifn>mwe multiply by 10
n
to get 10p+
a
n=10q, wherep, q∈N, so thata n= 10(q−p). Sinceq−p∈Zwhilea nis
one of the digits 0,1,...,9, it follows thata
n= 0, a contradiction. Therefore
n≤m, and a similar argument shows thatm≤n; thereforen=m.
Repeating the above argument withn=m, we obtain 10p+a
n=10q+b n,
so thata
n−bn= 10(q−p), whence it follows thata n=bn. If this argument
is repeated, we conclude thata
k=bkfork=1,...,n.
15. The problem here is that−2/7 is a negative number, so we write it as
−1+5/7. Since 5/7=.714285···with the block repeating, we write−2/7=
−1+.714285
16. 1/7=.142857···,the block repeats. 2/19 =.105263157894736842···,the
block repeats.
17. 1.25137···137···= 31253/24975,35.14653···653···= 3511139/99900.

CHAPTER 3
SEQUENCES
Most students will ﬁnd this chapter easier to understand than the preceding one
for two reasons: (i) they have a partial familiarity with the notions of a sequence
and its limit, and (ii) it is a bit clearer what one can use in proofs than it was for
the results in Chapter 2. However, since it is essential that the students develop
some technique, one should not try to go too fast.
Section 3.1
The main diﬃculty students have is mastering the notion oflimitof a sequence,
given in terms ofεandK(ε). Students should memorize the deﬁnition accurately.
The diﬀerent quantiﬁers in statements of the form “given any..., andthere
exists...” can beconfusing initially. We often use theK(ε) game as a device
to emphasize exactly how the quantities are related in proving statements about
limits. The facts that theε>0 comes ﬁrst and is arbitrary, and that the index
K(ε) depends on it (but is not unique) must be stressed.
The idea of deriving estimates is important and Theorem 3.1.10 is often used
as a means of establishing convergence of a sequence by squeezing|x
n−x|between
0 and a ﬁxed multiple of|a
n|.
A careful and detailed examination of the examples in 3.1.11 is very instruc-
tive. Although some of the arguments may seem a bit artiﬁcial, the particular
limits established there are useful for later work, so the results should be noted
and remembered.
Sample Assignment: Exercises 1, 2(a,c), 3(b,d), 5(b,d), 6(a,c), 8, 10, 14,
15, 16.
Partial Solutions:
1. (a) 0, 2, 0, 2, 0, (b) −1,1/2,−1/3,1/4,−1/5,
(c) 1/2, 1/6, 1/12, 1/20, 1/30, (d) 1/3, 1/6, 1/11, 1/18, 1/27.
2. (a) 2n+ 3, (b) (−1)
n+1
/2
n
, (c)n/(n+ 1), (d)n
2
.
3. (a) 1, 4, 13, 40, 121, (b) 2, 3/2, 17/12, 577/408, 665, 857/470, 832,
(c) 1, 2, 3, 5, 4, (d) 3, 5, 8, 13, 21.
4. Givenε>0, takeK(ε)≥|b|/εifb =0.
5. (a) We have 0<n/(n
2
+1)< n/n
2
=1/n. Givenε>0, letK(ε)≥1/ε.
(b) We have|2n/(n+1)−2|=2/(n+1)<2/n. Givenε>0, letK(ε)≥2/ε.
(c) We have|(3n+1)/(2n+5)−3/2|=13/(4n+ 10)<13/4n. Givenε>0,
letK(ε)≥13/4ε.
(d) We have|(n
2
−1)/(2n
2
+3)−1/2|=5/(4n
2
+6)<5/4n
2
≤5/4n. Given
ε>0, letK(ε)≥5/4ε.
17

18 Bartle and Sherbert
6. (a) 1/
√
n+7<1/
√
n, (b) |2n/(n+2)−2|=4/(n+2)<4/n,
(c)
√
n/(n+1)<1/
√
n, (d)|(−1)
n
n/(n
2
+1)|≤1/n.
7. (a) [1/ln(n+1)<ε]⇐⇒[ln(n+1)>1/ε]⇐⇒[n+1>e
1/ε
]. Givenε>0, let
K≥e
1/ε
−1.
(b) Ifε=1/2, thene
2
−1≈6.389, so we chooseK=7. Ifε=1/10, then
e
10
−1≈22,025.466, so we chooseK= 22,026.
8. Note that||x
n|−0|=|x n−0|. Consider ((−1)
n
).
9. 0<
√
xn<ε⇐⇒0<x n<ε
2
.
10. Letε:=x/2. IfM:=K(ε), thenn≥Mimplies that|x−x
n|<ε=x/2, which
implies thatx
n>x−x/2=x/2>0.
11.|1/n−1/(n+1)|=1/n(n+1)<1/n
2
≤1/n.
12. Multiply and divide by
√
n
2
+1+nto obtain
√
n
2
+1−n=1/(
√
n
2
+1+n)<
1/n.
13. Note thatn<3
n
so that 0<1/3
n
<1/n.
14. Letb:= 1/(1 +a) wherea>0. Since (1 +a)
n
>
1
2
n(n−1)a
2
, we have
0<nb
n
≤n/[
1
2
n(n−1)a
2
]≤2/[(n−1)a
2
]. Thus lim(nb
n
)=0.
15. Use the argument in 3.1.11(d). If (2n)
1/n
=1+k n, then show thatk
2
n
≤
2(2n−1)/n(n−1)<4/(n−1).
16. Ifn>3, then 0<n
2
/n!<n/(n−2)(n−1)<1/(n−3).
17.
2
n
n!
=
2·2·2·2···2
1·2·3·4···n
=2·1·
2
3
·
2
4
···
2
n
≤2·
2
3
·
2
3
···
2
3
=2
π
2
3
α
n−2
.
18. Ifε:=x/2, thenn>K(ε) implies that|x−x
n|<x/2, which is equivalent to
x/2<x
n<3x/2<2x.
Section 3.2
The results in this section, at least beginning with Theorem 3.2.3, are clearly
useful in calculating limits of sequences. They are also easy to remember. The
proofs of the basic theorems use techniques that will recur in later work, and so are
worth attention (but not memorization). It may be pointed out to the students
that the Ratio Test in 3.2.11 has the same hypothesis as the Ratio Test for the
convergence of series that they encountered in their calculus course. There are
additional results of this nature in the exercises.
Sample Assignment: Exercises 1, 3, 5, 7, 9, 10, 12, 13, 14.
Partial Solutions:
1. (a) lim(x
n) = 1. (b) Divergence.
(c)x
n≥n/2, so the sequence diverges.
(d) lim(x
n) = lim(2 + 1/(n
2
+1))=2.
2. (a)X:= (n),Y:= (−n)orX:= ((−1)
n
),Y:= ((−1)
n+1
). Many other exam-
ples are possible. (b)X=Y:= ((−1)
n
).

Chapter 3 — Sequences 19
3.Y=(X+Y)−X.
4. Ifz
n:=xnynand lim(x n)=x = 0, then ultimatelyx n = 0 so thaty n=zn/xn.
5. (a) (2
n
) is not bounded since 2
n
>nby Exercise 1.2.13.
(b) The sequence is not bounded.
6. (a) (lim(2 + 1/n))
2
=2
2
= 4, (b) 0, since|(−1)
n
/(n+ 2))|≤1/n,
(c) lim
π
1−1/
√
n
1+1/
√
n
α
=
1
1
= 1, (d) lim(1/n
1/2
+1/n
3/2
)=0+0=0.
7. If|b
n|≤B,B>0, andε>0, letKbe such that|a n|< ε/Bforn>K.To
apply Theorem 3.2.3, it is necessary that both (a
n) and (b n) converge, but a
bounded sequence may not be convergent.
8. In (3) the exponentkis ﬁxed, but in (1 + 1/n)
n
the exponent varies.
9. Sincey
n=
1
√
n+1+
√
n
, we have lim(y
n) = 0. Also we have
√
nyn=
√
n
√
n+1+
√
n
=
1
Σ
1+1/n+1
, so that lim(
√
nyn)=
1
2
.
10. (a) Multiply and divide by
√
4n
2
+n+2nto obtain 1/(
Σ
4+1/n+ 2) which
has limit 1/4.
(b) Multiply and divide by
√
n
2
+5n+nto obtain 5/(
Σ
1+5/n+ 1) which
has limit 5/2.
11. (a) (
√
3)
1/n
(n
1/n
)
1/4
converges to 1·1
1/4
=1.
(b) Show that (n+1)
1/ln(n+1)
=efor alln∈N.
12.
a(a/b)
n
+b
(a/b)
n
+1
has limit
0+b
0+1
=bsince 0< a/b <1.
13.
(n+a)(n+b)−n
2
Σ
(n+a)(n+b)+n
=
(a+b)n+ab
Σ
(n+a)(n+b)+n
·
1/n
1/n
=
a+b+ab/n
Σ
(1 +a/n)(1 +b/n)+1
→
a+b
2
.
14. (a) Since 1≤n
1/n
2
≤n
1/n
, the limit is 1.
(b) Since 1≤n!≤n
n
implies 1≤(n!)
1/n
2
≤n
1/n
, the limit is 1.
15. Show thatb≤z
n≤2
1/n
b.
16. (a)L=a, (b)L=b/2, (c)L=1/b, (d)L=8/9.
17. (a) (1/n), (b) (n).
18. If 1<r<L, letε:=L−r. Then there existsKsuch that|x
n+1/xn−L|<ε
forn>K. From this one getsx
n+1/xn>rforn>K.Ifn>K, then
x
n≥r
n−K
xK. Sincer>1, it follows that (x n) is not bounded.
19. (a) Converges to 0, (b) Diverges,
(c) Converges to 0, (d)n!/n
n
≤1/n.

20 Bartle and Sherbert
20. IfL<r<1 andε:=r−L, then there existsKsuch that|x
1/n
n
−L|<ε=
r−Lforn>K, which implies thatx
1/n
n
<rforn>K. Then 0<x n<r
n
for
n>K, and since 0<r<1, we have lim(r
n
) = 0. Hence lim(x n)=0.
21. (a) (l), (b) (n).
22. Yes. The hypothesis implies that lim(y
n−xn) = 0. Sincey n=(yn−xn)+x n,
it follows that lim(y
n) = lim(x n).
23. It follows from Exercise 2.2.18 thatu
n=
1
2
(xn+yn+|xn−yn|). Theorems 3.2.3
and 3.2.9 imply that lim(u
n)=
1
2
[lim(x n) + lim(y n)+|lim(x n)−lim(y n)|]=
max{lim(x
n),lim(y n)}. Similarly for lim(v n).
24. Since it follows from Exercises 2.2.18(b) and 2.2.19 that mid{a, b, c}=
min{max{a, b},max{b, c},max{c, a}}, this result follows from the preceding
exercise.
Section 3.3
The Monotone Convergence Theorem 3.3.2 is a very important (and natural)
result. It implies theexistenceof the limit of a bounded monotone sequence.
Although it does not give an easy way of calculating the limit, it does give some
estimates about its value. For example, if (x
n) is an increasing sequence with
upper boundb, then limitx
∗
must satisfyx n≤x
∗
≤bfor anyn∈N. If this is
not suﬃciently exact, takex
mform>nand look for a smaller boundb
ﬃ
for the
sequence.
Sample Assignment: Exercises 1, 2, 4, 5, 7, 9, 10.
Partial Solutions:
1. Note thatx
2=6<x 1. Also, ifx k+1<xk, thenx k+2=
1
2
xk+1+2<
1
2
xk+2=
x
k+1. By Induction, (x n) is a decreasing sequence. Also 0≤x n≤8 for all
n∈N. The limitx:= lim(x
n) satisﬁesx=
1
2
x+ 2, so thatx=4.
2. Show, by Induction, that 1<x
n≤2 forn≥2 and that (x n) is monotone.
In fact, (x
n) is decreasing, for ifx 1<x2, then we would have (x 1−1)
2
<
x
2
1
−2x 1+1=0. Sincex:= lim(x n) must satisfyx=2−1/x, we have
x=1.
3. Ifx
k≥2, thenx k+1:=1+
√
xk−1≥1+
√
2−1 = 2, sox n≥2 for alln∈N,
by Induction. Ifx
k+1≤xk, thenx k+2=1+
√
xk+1−1≤1+
√
xk−1=
x
k+1,so(x n) is decreasing. The limitx:= lim(x n) satisﬁesx=1+
√
x−1so
thatx=1 orx= 2. Sincex= 1 is impossible (why?), we havex=2.
4. Note thaty
1=1<
√
3=y 2, and ify n+1−yn>0, theny n+2−yn+1=
(y
n+1−yn)/(
√
2+y n+1+
√
2+y n)>0, so (y n) is increasing by Induction.
Alsoy
1<2 and ify n<2, theny n+1=
√
2+y n<
√
2+2=2,so(y n) is bounded
above. Therefore (y
n) converges to a numberywhich must satisfyy=
√
2+y,
whencey=2.

Chapter 3 — Sequences 21
5. We havey
2=
Σ
p+
√
p>
√
p=y 1. Alsoy n>yn−1implies thaty n+1=
√
p+y n>
√
p+y n−1=yn,so(y n) is increasing. An upper bound for (y n)
isB:=1+2
√
pbecausey 1≤Band ify n≤Btheny n+1<
√
p+B=
1+
√
p<B.Ify:= lim(y n), theny=
√
p+yso thaty=
1
2
(1 +
√
1+4p).
6. Show that the sequence is monotone. The positive root of the equation
z
2
−z−a=0isz
∗
:=
1
2
(1 +
√
1+4a). Show that if 0<z 1<z
∗
, thenz
2
1
−z1−
a<0 and the sequence increase toz
∗
.Ifz
∗
<z1, then the sequence decreases
toz
∗
.
7. Sincex
n>0 for alln∈N, we havex n+1=xn+1/x n>xn, so that (x n)is
increasing. Ifx
n≤bfor alln∈N, thenx n+1−xn=1/x n≥1/b >0 for alln.
But if lim(x
n) exists, then lim(x n+1−xn) = 0, a contradiction. Therefore
(x
n) diverges.
8. The sequence (a
n) is increasing and is bounded above byb 1,soξ:= lim(a n)
exists. Also (b
n) is decreasing and bounded below bya 1soη:= lim(b n)
exists. Sinceb
n−an≥0 for alln, we haveη−ξ≥0. Thusa n≤ξ≤η≤b n
for alln∈N.
9. Show that ifx
1,x2,...,xn−1have been chosen, then there existsx n∈Asuch
thatx
n>u−1/nandx n≥xkfork=1,2,...,n−1.
10. Sincey
n+1−yn=1/(2n+1)+1/(2n+2)−1/(n+1)=1(2n+ 1)(2n+2)>0,
it follows that (y
n) is increasing. Alsoy n=1/(n+1)+1/(n+2)+···+
1/2n<1/(n+1)+1/(n+1)+···+1/(n+1)=n/(n+1)<1, so that (y
n)
is bounded above. Thus (y
n) is convergent. (It can be show that its limit
is ln 2).
11. The sequence (x
n) is increasing. Alsox n<1+1/1·2+1/2·3+···+
1/(n−1)n= 1+(1−1/2)+(1/2−1/3)+···+(1/(n−1)−1/n)=2−1/n <2,
so (x
n) is bounded above and (x n) is convergent. (It can be shown that its
limit isπ
2
/6).
12. (a) (1 + 1/n)
n
(1+1/n)→e·1=e, (b) [(1 + 1/n)
n
]
2
→e
2
,
(c) [1 + 1/(n+ 1)]
n+1
/[1+1/(n+ 1)]→e/1=e,
(d) (1−1/n)
n
=[1+1/(n−1)]
−n
→e
−1
=1/e.
13. Note that ifn≥2, then 0≤s
n−
√
2≤s
2
n
−2.
14. Note that 0≤s
n−
√
5≤(s
2
n
−5)/
√5≤(s
2
n
−5)/2.
15.e
2=2.25,e 4=2.441 406,e 8=2.565 785,e 16=2.637 928.
16.e
50=2.691 588,e 100=2.704 814,e 1000=2.716 924.
Section 3.4
The notion of a subsequence is extremely important and will be used often. It
must be emphasized to students that a subsequence is not simply a collection of
terms, but an ordered selection that is a sequence in its own right. Moreover, the

22 Bartle and Sherbert
order is inherited from the order of the given sequence. The distinction between
a sequence and a set is crucial here.
The Bolzano-Weierstrass Theorem 3.4.8 is a fundamental result whose impor-
tance cannot be over-emphasized. It will be used as a crucial tool in establishing
the basic properties of continuous functions in Chapter 5.
Sample Assignment: Exercises 1, 2, 3, 5, 6, 9, 12.
Partial Solutions:
1. Letx
2n−1:= 2n−1,x 2n:= 1/2n; that is (x n)=(1,1/2,3,1/4,5,1/6,...).
2. Ifx
n:=c
1/n
, where 0<c<1, then (x n) is increasing and bounded, so it has a
limitx. Sincex
2n=
√
xn, the limit satisﬁesx=
√
x,sox=0 orx= 1. Since
x= 0 is impossible (why?), we havex=1.
3. Sincex
n≥1 for alln∈N,L>0. Further, we havex n=1/x n−1+1⇒
L=1/L+1⇒L
2
−L−1=0⇒L=
1
2
(1 +
√
5).
4. (a)x
2n→0 andx 2n+1→2.
(b)x
8n= 0 andx 8n+1= sin(π/4)=1/
√
2 for alln∈N.
5. If|x
n−z|<εforn≥K 1and|y n−z|<εforn≥K 2, letK:= sup{2K 1−
1,2K
2}. Then|z n−z|<εforn≥K.
6. (a)x
n+1<xn⇐⇒(n+1)
1/(n+1)
<n
1/n
⇐⇒(n+1)
n
<n
n+1
=n
n
·n⇐⇒
(1+1/n)
n
<n.
(b) Ifx:= lim(x
n), then
x= lim(x
2n) = lim((2n)
1/2n
) = lim((2
1/n
n
1/n
)
1/2
)=x
1/2
,
so thatx=0 orx= 1. Sincex
n≥1 for alln, we havex=1.
7. (a) (1 + 1/n
2
)
n
2
→e,
(b) (1 + 1/2n)
n
=((1+1/2n)
2n
)
1/2
→e
1/2
,
(c) (1 + 1/n
2
)
2n
2
→e
2
.
(d) (1 + 2/n)
n
=(1+1/(n+ 1))
n
·(1+1/n)
n
→e·e=e
2
.
8. (a) (3n)
1/2n
= ((3n)
1/3n
)
3/2
→1
3/2
=1,
(b) (1 + 1/2n)
3n
=((1+1/2n)
2n
)
3/2
→e
3/2
.
9. If (x
n) does not converge to 0, then there existsε 0>0 and a subsequence
(x
nk) with|x nk|>ε0for allk∈N.
10. Choosem
1such thatS≤s m1<S+ 1. Now choosek 1such thatk 1≥m1and
s
m1−1<x k1
≤sm1.Ifm 1<m2<···<m n−1andk 1<k2<···<k n−1have
been selected, choosem
n>mn−1such thatS≤s mn<S+1/n. Now choose
k
n≥mnandk n>kn−1such thats mn−1/n<x kn
≤smn. Then (x kn
)isa
subsequence of (x
n) and|x kn
−S|≤1/n.

Chapter 3 — Sequences 23
11. Show that lim((−1)
n
xn)=0.
12. Choosen
1≥1 so that|x n1|>1, then choosen 2>n1so that|x n2|>2, and,
in general, choosen
k>nk−1so that|x nk|>k.
13. (x
2n−1)=(−1,−1/3,−1/5,...).
14. Choosen
1≥1 so thatx n1≥s−1, then choosen 2>n1so thatx n2>s−1/2,
and, in general, choosen
k>nk−1so thatx nk>s−1/k.
15. Suppose that the subsequence (x
nk) converges tox. Givenn∈Nthere exists
Ksuch that ifk≥Kthenn
k≥n, so thatx nk∈Ink⊆In=[an,bn] for all
k≥K. By 3.2.6 we conclude thatx= lim(x
nk) belongs toI nfor arbitrary
n∈N.
16. For example,X= (1, 1/2, 3, 1/4, 5, 1/6,...).
17. lim sup(x
n)=1,sup{x n}=2,lim inf(x n)=0,inf{x n}=−1.
18. Ifx= lim(x
n) andε>0 is given, then there existsNsuch thatx−ε<
x
n<x+εforn≥N. The second inequality implies lim sup(x n)≤x+ε
and the ﬁrst inequality implies lim inf(x
n)≥x−ε. Then 0≤lim sup(x n)−
lim inf(x
n)≤2ε. Sinceε>0 is arbitrary, equality follows. Conversely, if
x= lim inf(x
n) = lim sup(x n), then there existsN 1such thatx n<x+εfor
n>N
1, andN 2such thatx−ε<x nforn≥N 2. Now takeNto be the larger
ofN
1andN 2.
19. Ifv>lim sup(x
n) andu>lim sup(y n), then there are at most ﬁnitely many
nsuch thatx
n>vand at most ﬁnitely manynsuch thaty n>v. Therefore,
there are at most ﬁnitely manynsuch thatx
n+yn>v+u, which implies
lim sup(x
n+yn)≤v+u. This proves the stated inequality. For an example
of strict inequality, one can takex
n=(−1)
n
andy n=(−1)
n+1
.
Section 3.5
At ﬁrst, students may encounter a little diﬃculty in working with Cauchy
sequences. It should be emphasized that in proving that a sequence (x
n)isa
Cauchy sequence, the indicesn, min Deﬁnition 3.5.1 are completely indepen-
dent of one another (however, one can always assume thatm>n). On the other
hand, to prove that a sequence isnota Cauchy sequence, a particular relationship
betweennandmcan be assumed in the process of showing that the deﬁnition is
violated.
The signiﬁcance of Cauchy criteria for convergence is not immediately appar-
ent to students. Its true role in analysis will be slowly revealed by its use in
subsequent chapters.
We have included the discussion of contractive sequences to illustrate just
one way in which Cauchy sequences can arise.

24 Bartle and Sherbert
Sample Assignment: Exercises 1, 2, 3, 5, 7, 9, 10.
Partial Solutions:
1. For example, ((−1)
n
).
2. (a) Ifm>n, then|(1+1/m)−(1+1/n)|<2/n.
(b) 0<1/(n+ 1)! +···+1/m!<1/2
n
, since 2
k
<k! fork≥4.
3. (a) Note that|(−1)
n
−(−1)
n+1
|= 2 for alln∈N.
(b) Takem=2n,sox
m−xn=x2n−xn≥1 for alln.
(c) Takem=2n,sox
m−xn=x2n−xn=ln2n−lnn= ln 2 for alln.
4. Use|x
mym−xnyn|≤|y m||xm−xn|+|x n||ym−yn|and the fact that Cauchy
sequences are bounded.
5. lim(
√
n+1−
√
n) = lim
π
1
√
n+1+
√
n
α
= 0. However, ifm=4n, then
√
4n−
√
n=
√
nfor alln.
6. Letx
n:=1+1/2+···+1/n, which is not a Cauchy sequence. (Why?)
However, for anyp∈N, then 0<x
n+p−xn=1/(n+1)+···+1/(n+p)≤
p/(n+ 1), which has limit 0.
7. Ifx
n,xmare integers and|x m−xn|<1, thenx n=xm.
8. Letu:= sup{x
n:n∈N}.Ifε>0, letHbe such thatu−ε<x H≤u.If
m≥n≥H, thenu−ε<x
n≤xm≤uso that|x m−xn|<ε.
9. Ifm>n, then|x
m−xn|<r
n
+r
n+1
+···+r
m−1
≤r
n
/(1−r), which con-
verges to 0 since 0<r<1.
10. IfL:=x
2−x1, then|x n+1−xn|=L/2
n−1
, whence it follows that (x n)isa
Cauchy sequence. To ﬁnd the limit, show thatx
2n+1=x1+L/2+L/2
3
+
L/2
5
+···+L/2
2n−1
, whence lim(x n)=x 1+(2/3)L=(1/3)x 1+(2/3)x 2.
11. Note that|y
n−yn+1|=(2/3)|y n−yn−1|. Sincey 2>y1, the limit isy=y 1+
(3/5)(y
2−y1)=(2/5)y 1+(3/5)y 2.
12. Show that|x
n+1−xn|<
1
4
|xn−xn−1|. The limit is
√
2−1.
13. Note thatx
n≥2 for alln, so that|x n+1−xn|=|1/x n−1/x n−1|=
|x
n−xn−1|/xnxn−1≤
1
4
|xn−xn−1|. The limit is 1 +
√
2.
14. Letx
n+1=(x
3
n
+1)/5 andx 1:= 0. Four iterations giver=0.201 64 to 5
decimal places.
Section 3.6 This section can be omitted on a ﬁrst reading. However, it is short, relatively easy,
and prepares the way for Section 4.3. One must frequently emphasize that∞and
−∞arenotreal numbers, but merely convenient abbreviations. While there is
no reason to expect that one can manipulate with properly divergent sequences
as one does in Theorem 3.2.3, there are some results in this direction.

Chapter 3 — Sequences 25
Sample Assignment: Exercises 1, 2, 3, 5, 8, 9.
Partial Solutions:
1. If the set{x
n:n∈N}is not bounded above, choosen k+1>nksuch that
x
nk≥kfork∈N.
2. (a)x
n:=
√
n, yn:=n, (b) x n:=n, yn:=
√
n.
3. Note that|x
n−0|<εif and only if 1/x n>1/ε.
4. (a) [
√
n>a]⇐⇒[n>a
2
], (b)
√
n+1>
√
n,
(c)
√
n−1≥
Σ
n/2 whenn≥2, (d)n/
√
n+1≥
√
n/2.
5. No. As in Example 3.4.6(c), there is a subsequence (n
k) withn ksin(n k)>
1
2
nk,
and there is a subsequence (m
k) withm ksin(m k)<−
1
2
mk.
6. If (y
n) does not converge to 0, there existsc>0 and a subsequence (y nk) with
|y
nk|≥c. Hence|x nk|=|x nkynk/ynk|is bounded, contradicting the fact that
(x
n) is properly divergent.
7. (a) There existsN
1such that ifn>N 1, then 0<x n<yn. If lim(x n)=∞
then lim(y
n)=∞.
(b) Suppose that|y
n|≤Mfor someM>0. Givenε>0 there existsN ε
such that ifn≥N εthen 0<x n/yn≤ε/M. Therefore|x n|≤(ε/M)y n≤εfor
n≥N
ε.
8. (a)n<(n
2
+2)
1/2
.
(b) Since
√
n≤n, then
√
n/(n
2
+1)≤n/(n
2
+1)<1/n.
(c) Sincen<(n
2
+1)
1/2
, thenn
1/2
<(n
2
+1)
1/2
/n
1/2
.
(d) If the sequence were convergent, the subsequence corresponding tor
k=k
2
would converge, contrary to Example 3.4.6(c).
9. (a) Sincex
n/yn→∞, there existsK 1such that ifn≥K 1, thenx n≥yn.Now
apply Theorem 3.6.4(a).
(b) Let 0<x
n<M.If(y n) does not converge to 0, there existε 0>0 and a
subsequence (y
nk) such thatε 0<ynk. Since lim(x n/yn)=∞, there existsK
such that ifk>K, thenM/ε
0<xnk/ynk, which is a contradiction.
10. Apply Theorem 3.6.5.
Section 3.7
This section gives a brief introduction to inﬁnite series, a topic that will be
discussed further in Chapter 9. However, since the basic results are merely a
reformulation of the material in Sections 3.1–3.6, it is useful to treat this section
here — especially, if there is a possibility that one might not be able to cover
Chapter 9 in class.
It must be made clear to the students that there is a signiﬁcant diﬀerence
between a “sequence” of numbers and a “series” of numbers. Indeed, a series is
a special kind of sequence, where the terms are obtained by adding terms in a

26 Bartle and Sherbert
given sequence. For a series to be convergent, the given terms must approach
0 “suﬃciently fast”. Unfortunately there is no clear demarcation line between
the convergent and the divergent series. Thus it is especially important for the
students to acquire a collection of series that are known to be convergent (or
divergent), so that these known series can be used for the purpose of comparison.
The speciﬁc series that are discussed in this section are particularly useful in this
connection.
Although much of the material in this section will be somewhat familiar to the
students, most of them will not have heard of the Cauchy Condensation Criterion
(Exercise 15), which is a very powerful testwhen it applies.
Sample Assignment: Exercises 1, 2, 3(a,b), 4, 8, 12, 15, 16, 17.
Partial Solutions:
1. Show that the partial sums of
β
b
nform a subsequence of the partial sums
of
β
a
n.
2. Ifa
n=bnforn>K, show that the partial sumss nof
β
a nandt nof
β
b n
satisfys n−tn=sK−tKfor alln>K.
3. (a) Note that 1/(n+ 1)(n+2)=1/(n+1)−1/(n+ 2), so the series is tele-
scoping and converges to 1.
(b) 1/(α+n)(α+n+1)=1/(α+n)−1/(α+n+ 1).
(c) 1/n(n+ 1)(n+2) = 1/2n−1/(n+1)+1/2(n+ 2), so that
β
N
1
=
1/4−1/2(N+1)+1/2(N+ 2).
4. Ifs
n:=
β
n
1
xkandt n:=
β
n
1
yk, thens n+tn=
β
n
1
(xk+yk).
5. No. Letz
k:=xk+ykso thaty k=zk+(−1)x k.If
β
(−1)x kand
β
z kare
convergent, then
β
y
kis convergent.
6. (a) (2/7)
2
[1/(1−2/7)]=4/35. (b) (1/9)[1/(1−1/9)]=1/8.
7.r
2
(1 +r
2
+(r
2
)
2
+···)=r
2
/(1−r
2
)
8.S=ε+ε
2
+ε
3
+···=ε/(1−ε).
S=1/9=0.111···ifε=0.1, andS=1/99=0.0101···ifε=0.01.
9. (a) The sequence (cosn) does not converge to 0.
(b) Since|(cosn)/n
2
|≤1/n
2
, the convergence of
β
(cosn)/n
2
follows from
Example 3.7.6(c) and Theorem 3.7.7.
10. Note that the “even” sequence (s
2n) is decreasing, and the “odd” sequence
(s
2n+1) is increasing and−1≤s n≤0. Moreover 0≤s 2n−s2n+1=
1/
√
2n+1.
11. If convergent, thena
n→0, so there existsM>0 such that 0<a n≤M,
whence 0<a
2
n
≤Ma n, and the Comparison Test 3.7.7 applies.
12.
β
1/n
2
is convergent, but
β
1/nis not.
13. Recall that ifa, b≥0 then 2
√
ab≤a+b,so
√
anan+1≤(an+an+1)/2. Now
apply the Comparison Test 3.7.7.

Chapter 3 — Sequences 27
14. Show thatb
k≥a1/kfork∈N, whenceb 1+···+b n≥a1(1 +···+1/n).
15. Evidently 2a(4)≤a(3) +a(4)≤2a(2) and 2
2
a(8)≤a(5) +···+a(8)≤2
2
a(4),
etc. The stated inequality follows by addition. Now apply the Comparison
Test 3.7.7.
16. Clearlyn
p
<(n+1)
p
ifp>0, so that the terms in the series are decreasing.
Since 2
n
·(1/2
pn
)=(1/2
p−1
)
n
, the Cauchy Condensation Test asserts that the
convergence of thep-series is the same as that of the geometric series with
ratio 1/2
p−1
, which is<1 whenp>1 and is≥1 whenp≤1.
17. (a) The terms are decreasing and 2
n
/2
n
ln(2
n
)=1/(nln 2). Since the har-
monic series
β
1/ndiverges, so does
β
1/(nlnn).
(b) 2
n
/2
n
(ln 2
n
)(ln ln 2
n
)=1/(nln 2)(lnn(ln 2)). Now use the Limit Compa-
rison Test 3.7.8 and part (a).
18. (a) The terms are decreasing and 2
n
/2
n
(ln 2
n
)
c
=(1/n
c
)·(1/ln 2)
c
. Now use
the fact that
β
(1/n
c
) converges whenc>1.
(b) Since ln(n/2)<ln(nln 2), we have 1/(ln(nlnn))
c
<1/(ln(n/2))
c
.Now
apply (a).

CHAPTER 4
LIMITS
In this chapter we begin the study of functions of a real variable. This and
the next chapter are the most important ones in the book, since all subsequent
material depends on the results in them. In Section 4.1 the concept of a limit of a
function at a point is introduced, and in Section 4.2 the basic properties of limits
are established. Both of these sections are necessary preparation for Chapter 5.
However, Section 4.3 can be omitted on a ﬁrst reading, if time is short.
Examples are a vital part of real analysis. Although certain examples do
not need to be discussed in detail, we advise that the students be urged to study
them carefully. One way of encouraging this is to ask for examples of various
phenomena on examinations.
Section 4.1
Attention should be called to the close parallel between Section 3.1 and this sec-
tion. It should be noted that hereδ(ε) plays the same role asK(ε) did in Section
3.1. The proof of the Sequential Convergence Theorem 4.1.8 is instructive and
the result is important. As a rule of thumb, theε-δformulation of the limit is
used to establish a limit, while sequences are more often used to (i) evaluate a
limit, or (ii) prove that a limit fails to exist.
Sample Assignment: Exercises 1, 3, 6, 8, 9, 10(b,d), 11(a), 12(a,c), 15.
Partial Solutions:
1. (a–c) If|x−1|≤1,then|x+1|≤3 so that|x
2
−1|≤3|x−1|. Thus,|x−1|<
1/6 assures that|x
2
−1|<1/2, etc.
(d) Sincex
3
−1=(x−1)(x
2
+x+1),if|x−1|<1, then 0<x<2 and so
|x
3
−1|≤7|x−1|.
2. (a) Since|
√
x−2|=|x−4|/(
√
x+2)≤
1
2
|x−4|,then|x−4|<1 implies that
|
√
x−2|<
1
2
.
(b) If|x−4|<2×10
−2
=.02, then|
√
x−2|<.01.
3. Apply the deﬁnition of the limit.
4. If lim
y→c
f(y)=L, then for anyε>0 there existsδ>0 such that if 0<|y−c|<δ,
then|f(y)−L|<ε. Now letx:=y−cso thaty=x+c,to conclude that
lim
x→0
f(x+c)=L.
5. If 0<x<a, then 0<x+c<a+c<2a,so that|x
2
−c
2
|=|x+c||x−c|≤
2a|x−c|.Givenε>0,takeδ:=ε/2a.
6. Takeδ:=ε/K.
28

Chapter 4 — Limits 29
7. Letb:=|c|+1.If|x|<b,then|x
2
+cx+c
2
|≤3b
2
.Hence|x
3
−c
3
|≤
(3b
2
)|x−c|for|x|<b.
8. Note that
√
x−
√
c=(
√
x−
√
c)(
√
x+
√
c)/(
√
x+
√
c)=(x−c)/(
√
x+
√
c).
Hence, ifc =0,we have|
√
x−
√
c|≤(1/
√
c)|x−c|,so that we can takeδ:=
ε
√
c.Ifc= 0, we can takeδ:=ε
2
,so that if 0<x<δ, then|
√
x−0|<ε.
9. (a) If|x−2|<1/2,thenx>3/2,sox−1>1/2 whence 0<1/(x−1)<2
and so|1/(1−x)+1|=|(x−2)/(x−1)|≤2|x−2|.Thus we can take
δ:= inf{1/2,ε/2}.
(b) If|x−1|<1,thenx+1>1,so 1/(x+1)<1 whence|x/(1 +x)−1/2|=
|x−1|/(2|x+1|)≤|x−1|/2≤|x−1|.Thus we may takeδ:= inf{1,ε}.
(c) Ifx =0,then|x
2
/|x|−0|=|x|.Takeδ:=ε.
(d) If|x−1|<1, then|2x−1|<3 and 1/|x+1|<1,so that|(x
2
−x+1)/
(x+1)−1/2|=|2x−1||x−1|/(2|x+1|)≤(3/2)|x−1|,so we may takeδ:=
inf{1,2ε/3}.
10. (a) If|x−2|<1, then|x
2
+4x−12|=|x+6||x−2|<9|x−2|.We may take
δ:= inf{1,ε/9}.
(b) If|x+1|<1/4,then−5/4<x<−3/4 so that 1/2<2x+3<3/2,and thus
0<1/(2x+3)<2.Then|(x+5)/(2x+3)−4|=7|x+1|/|2x+3|<14|x+1|,
so that we may takeδ:= inf{1/4,ε/14}.
11. (a) If|x−3|<1/2, thenx>5/2, so 4x−9>1 and then 1/|4x−9|<1. Then

2x+3
4x−9
−3

=

10x−30
4x−9

≤10|x−3|. Thus we takeδ= inf{1/2,ε/10}.
(b) If|x−6|<1, thenx+1<8, andx+3>8, so that (x+1)/(x+3)<1.
Then

x
2
−3x
x+3
−2

=

x+1
x+3

|x−6|≤|x−6|.Thus we takeδ= inf{1,ε}.
12. (a) Letx
n:= 1/n. (b) Let x n:= 1/n
2
.
(c) Letx
n:= 1/nandy n:=−1/n.
(d) Letx
n:= 1/
√
nπandy n:= 1/
Σ
π/2+2πn.
13. If|f(y)−L|<εfor|y|<δ, then|g(x)−L|<εfor 0<|x|<δ/a.
14. (a) Givenε>0, chooseδ>0 such that 0<|x−c|<δimplies (f(x))
2
<ε
2
.
(b) Iff(x) := sgn(x), then lim
x→0
(f(x))
2
= 1, but lim
x→0
f(x) does not exist.
15. (a) Since|f(x)−0|≤|x|, we can takeδ:=εto show that lim
x→0
f(x)=0.
(b) Ifc = 0 is rational, let (x
n) be a sequence of irrational numbers that
converges toc; thenf(c)=c = 0 = lim(f(x
n)). Ifcis irrational, let (r n)
be a sequence of rational numbers that converges toc; thenf(c)=0 =c=
lim(f(r
n)).
16. SinceIis an open interval containingc, there existsa>0 such that the
a-neighborhoodV
a(c)⊆I.Forε>0, ifδ>0 is chosen so thatδ≤a, then it
will apply to bothfandf
1.
17. The restriction of sgn to [0, 1] has a limit at 0.

30 Bartle and Sherbert
Section 4.2
Note the close parallel between this section and Section 3.2. While the proofs
should be read carefully, the main interest here is in the application of the theorems
to the calculation of limits.
Sample Assignment: Exercises 1, 2, 4, 5, 9, 11, 12.
Partial Solutions:
1. (a) 15, (b)−3, (c) 1/12, (d) 1/2.
2. (a) The limit is 1.
(b) Since (x
2
−4)/(x−2) =x+ 2 forx = 2, the limit is 4. Note that Theorem
4.2.4(b) cannot be applied here.
(c) The quotient equalsx+ 2 forx = 0. Hence the limit is 2.
(d) The quotient equals 1/(
√
x+ 1) forx =1.The limit is 1/2.
3. Multiply the numerator and denominator by
√
1+2x+
√
1+3x. The limit
is−1/2.
4. Ifx
n:= 1/2πnforn∈N, then cos(1/x n)=1.Also, ify n:= 1/(2πn+π/2) for
n∈N, then cos(1/y
n) = 0. Hence cos(1/x) does not have a limit asx→0.
Since|xcos(1/x)|≤|x|, the Squeeze Theorem 4.2.7 applies.
5. If|f(x)|≤Mforx∈V
δ(c), then|f(x)g(x)−0|≤M|g(x)−0|forx∈V δ(c).
6. Givenε>0, chooseδ
1>0 so that if 0<|x−c|<δ 1,x∈A, then|f(x)−L|<
ε/2. Chooseδ
2>0 so that if 0<|x−c|<δ 2,x∈A, then|g(x)−M|<ε/2.
Takeδ:= inf{δ
1,δ2}.Ifx∈Asatisﬁes 0<|x−c|<δ, then|(f(x)+g(x))−
(L+M)|≤|f(x)−L|+|g(x)−M|<ε/2+ε/2=ε.
7. Let (x
n) be any sequence inA\{c}that converges toc. Then (f(x n)) con-
verges toLand (h(x
n)) converges toH. By 3.23(b), (f(x n)/h(x n)) converges
toL/H. Since (x
n) is an arbitrary sequence inA\{c}, it follows from 4.1.8
that lim
x→c
f/h=L/H.
8. If|x|≤1,k∈N, then|x
k
|=|x|
k
≤1, whence−x
2
≤x
k+2
≤x
2
. Thus, ifn≥2,
we have|x
n
−0|≤|x
2
−0|for|x|≤1.Consequently lim
x→0
x
n
=0.
9. (a) Note thatg(x)=(f+g)(x)−f(x).
(b) No; for example, takef(x)=x
2
andg(x):=1/xforx>0.
10. Letf(x):=1 ifxis rational andf(x):=0 ifxis irrational, and let
g(x):=1−f(x). Thenf(x)+g(x) = 1 for allx∈R, so that lim
x→0
(f+g)=1,
andf(x)g(x) = 0 for allx∈R, so that lim
x→0
fg=0.
11. (a) No limit, (b) 0, (c) No limit, (d) 0.
12. Sincef((k+1)y)=f(ky+y)=f(ky)+f(y), an induction argument shows
thatf(ny)=nf(y) for alln∈N,y∈R. If we substitutey:= 1/n, we get
f(1/n)=f(1)/n, whenceL= lim
x→0
f(x) = lim(f(1/n)) = 0. Sincef(x)−f(c)=

Chapter 4 — Limits 31
f(x−c), we infer that lim
x→c
(f(x)−f(c)) = lim
x→c
f(x−c) = lim
z→0
f(z) = 0, so that
lim
x→c
f(x)=f(c).
13. (a)g(f(x)) =g(x+1)=2 ifx = 0, so that lim
x→0
g(f(x)) = 2, butg(lim
x→0
f(x)) =
g(f(0)) =g(1) = 0. Not equal.
(b)f(g(x)) =g(x)+1=3ifx = 1, so that lim
x→1
f(g(x)) = 3, andf(lim
x→1
g(x)) =
f(2) = 3. Equal.
14. If lim
x→c
f(x)=L, then||f(x)|−|L||≤|f(x)−L|implies that lim
x→c
|f(x)|=|L|.
15. This follows from Theorem 3.2.10 and the Sequential Criterion 4.1.8. Alter-
natively, anε-δproof can be given.
Section 4.3
This section can play the role of reinforcing the notion of the limit, since it provides
several extensions of this concept. However, the results obtained here are used in
only a few places later, so that it is easy to omit this section on a ﬁrst reading.
In fact, one-sided limits are used only once or twice in subsequent chapters.
In any case, we advise that the discussion of this section be quite brief.
Indeed, it is quite reasonable to give a short introduction to it in a class, and
leave it to the students to return to it later, when needed.
Sample Assignment: Exercises 2, 3, 4, 5(a,c,e,g), 8, 9.
Partial Solutions:
1. Modify the proof of Theorem 4.1.8 appropriately. Note that 0<|x−c|<δis
replaced by 0<x−c<δsincex>c.
2. Letf(x) := sin(1/x) forx<0 andf(x):=0 forx>0.
3. Givenα>0, if 0<x<1/α
2
, then
√
x<1/α, and sof(x)>α. Sinceαis
arbitrary, lim
x→0+
x/(x−1) =∞.
4. Ifα>0, thenf(x)>αif and only if|1/f(x)−0|<1/α.
5. (a) Ifα>1 and 1<x<α/(α−1), thenα<x/(x−1), hence we have
lim
x→1+
x/(x−1) =∞.
(b) The right-hand limit is∞; the left-hand limit is−∞.
(c) Since (x+2)/
√
x>2/
√
x, the limit is∞.
(d) Since (x+2)/
√
x>
√
x, the limit is∞.
(e) Ifx>0, then 1/
√
x<(
√
x+1)/x, so the right-hand limit is∞. What is
the left-hand limit?
(f) 0. (g) 1. (h) −1.
6. Modify the proof of Theorem 4.3.2 (using Deﬁnition 4.3.10). Note that
0<x−c<δ(ε) is replaced byx>K(ε).
7. Use Theorem 4.3.11.

32 Bartle and Sherbert
8. Note that|f(x)−L|<εforx>Kif and only if|f(1/z)−L|<εfor
0<z<1/K.
9. There existsα>0 such that|xf(x)−L|<1 wheneverx>α. Hence|f(x)|<
(|L|+1)/xforx>α.
10. Modify the proof of Theorem 4.3.11 (using Deﬁnition 4.3.13). Note that
|f(x)−L|<εis replaced byf(x)>α[respectively,f(x)<α].
11. Letα>0 be arbitrary and letβ>(2/L)α. There existsδ
1>0 such that
if 0<|x−c|<δ
1thenf(x)>L/2, and there existsδ 2>0 such that if
0<|x−c|<δ
2, theng(x)>β.Ifδ 3:= inf{δ 1,δ2}, and if 0<|x−c|<δ 3then
f(x)g(x)>(L/2)β>α. Sinceαis arbitrary, then lim
x→c
fg=∞. Letc= 0 and
letf(x):=|x|andg(x):=1/|x|forx =0.
12. Letf(x)=g(x):=x(or letf(x):=xandg(x):=x+1/x).
No. Ifh(x):=f(x)−g(x), thenf(x)/g(x)=1+h(x)/g(x)→1.
13. Suppose that|f(x)−L|<εforx>K, and thatg(y)>Kfory>H. Then
|f◦g(y)−L|<εfory>H.

CHAPTER 5
CONTINUOUS FUNCTIONS
This chapter can be considered to be the heart of the course. We now use all the
machinery that has been developed to this point in order to study the most impor-
tant class of functions in analysis, namely,continuous functions. In Section 5.3,
the fundamental properties of continuous functions are proved, and this section is
the most important of this chapter. Suﬃcient time should be spent on it to allow
adequate study of the proofs and examples. Section 5.4 on uniform continuity is
also an important section. Section 5.5 contains a diﬀerent approach to the basic
theorems in Sections 5.3 and 5.4, using the idea of a “gauge”.
The results on monotone functions in Section 5.6 are interesting, but they
are not central to this course and these results will not be used often in later parts
of this book.
Section 5.1
This important section is absolutely basic to everything that will follow. Every
eﬀort should be made to have the students master the notions presented here.
They should memorize the deﬁnition of continuity and its various equivalents,
and they should study the examples very carefully.
Sample Assignment: Exercises 1, 3, 4(a,b), 5, 7, 11, 12, 13.
Partial Solutions:
3. We will establish the continuity ofhatb. Sincefis continuous atb, given
ε>0 there existsδ
1>0 such that ifb−δ 1<x<b, then|f(x)−f(b)|<ε. Sim-
ilarly, there existsδ
2>0 such that ifb<x<b+δ 2, then|g(x)−g(b)|<ε. Let
δ:= inf{δ
1,δ2}so that|h(x)−h(b)|<εfor|x−b|<δ, whencehis continuous
atb.
4. (a) Continuous ifx =0,±1,±2,...,
(b) Continuous ifx =±1,±2,...,
(c) Continuous if sinx =0,1,
(d) Continuous ifx =0,±1,±1/2,....
5. Yes. Deﬁnef(2) := lim
x→2
f(x)=5.
6. Givenε>0, chooseδ>0 such that ifx∈V
δ(c)∩A, then|f(x)−f(c)|<ε/2.
Then ify∈V
δ(c)∩A, we have|f(y)−f(x)|≤|f(x)−f(c)|+|f(c)−f(y)|<
ε/2+ε/2=ε.
7. Letε:=f(c)/2, and letδ>0 be such that if|x−c|<δ, then|f(x)−f(c)|<ε,
which implies thatf(x)>f(c)−ε=f(c)/2>0.
8. Sincefis continuous atx,wehavef(x) = lim(f(x
n))=0. Thusx∈S.
33

34 Bartle and Sherbert
9. (a) If|f(x)−f(c)|<εforx∈V
δ∩B, then|g(x)−g(c)|=|f(x)−f(c)|<εfor
x∈V
δ(c)∩A.
(b) Letf= sgn onB:= [0,1],g= sgn onA:= (0,1] andc=0.
10. Note that

|x|−|c|

≤|x−c|.
11. Letc∈Rbe given and letε>0. If|x−c|< ε/K, then|f(x)−f(c)|≤
K|x−c|<K(ε/K)=ε.
12. Ifxis irrational, then by the Density Theorem 2.4.8 there exists a sequence
(r
n) of rational numbers that converges tox. Thenf(x) = lim(f(r n))=0.
13. Since|g(x)−6|≤sup{|2x−6|,|x−3|}=2|x−3|, thengis continuous at
x=3. Ifc = 3, let (x
n) be a sequence of rational numbers converging to
cand let (y
n) be a sequence of irrational numbers converging toc. Then
lim(g(x
n))=2c =c+ 3 = lim(g(y n)), sogis not continuous atc.
14. Letc∈A.Ifkis continuous atc, it follows from 4.2.2 thatkis bounded
on some neighborhood (c−δ, c+δ). Letm∈Nbe given; then there exists
a prime numberpsuch that 1/p<δandp≥m. (Why?) There must be at
least one rational numberq/pwithc−δ<q/p<c+δ; otherwise there exists
an integerq
0such thatq 0/p≤c−δandc+δ≤(q 0+1)/p, which implies that
2δ≤1/p, a contradiction. We conclude thatk(x)=p≥mfor at least one
pointx∈(c−δ, c+δ). But this is a contradiction.
15. LetI
n:= (0,1/n] forn∈N. Show that (supf(I n)) is a decreasing sequence
and (inff(I
n)) is an increasing sequence. If lim(supf(I n)) = lim(inff(I n)),
then lim
x→0
fexists. Letx n,yn∈Inbe such thatf(x n)>supf(I n)−1/nand
f(y
n)<inff(I n)+1/n.
Section 5.2
Note the similarity of this section with Sections 4.2 and 3.2. However, Theorem
5.2.6 concerning composite functions is a new result, and animportantone. Its
importance may be suggested by the fact, noted in 5.2.8, that it implies several
of the earlier results.
The signiﬁcance of this section should be clear: it enables us to establish the
continuity of many functions.
Sample Assignment: Exercises 1, 3, 5, 6, 10, 12, 13.
Partial Solutions:
1. (a) Continuous onR, (b) Continuous for x≥0,
(c) Continuous forx = 0, (d) Continuous onR.
2. Use 5.2.1(a) and Induction; or, use 5.2.8 withg(x):=x
n
.
3. Letfbe the Dirichlet discontinuous function (Example 5.1.6(g)) and let
g(x):=1−f(x).
4. Continuous at every noninteger.

Chapter 5 — Continuous Functions 35
5. The functiongis not continuous at 1 =f(0).
6. Givenε>0, there existsδ
1>0 such that if|y−b|<δ 1, then|g(y)−g(b)|<ε.
Further, there existsδ>0 such that if 0<|x−c|<δ, then|f(x)−b|<δ
1.
Hence, if 0<|x−c|<δ, then we have|(g◦f)(x)−g(b)|<ε, so that
lim
x→c
(g◦f)(x)=g(b).
7. Letf(x):=1 ifxis rational, andf(x):=−1ifxis irrational.
8. Yes. Givenx∈R, let (r
n) be a sequence of rational numbers withr n→x.
9. Show that an arbitrary real number is the limit of a sequence of numbers of
the formm/2
n
, wherem∈Z,n∈N.
10. Ifc∈P, thenf(c)>0. Now apply Theorem 4.2.9.
11. Ifh(x):=f(x)−g(x), thenhis continuous andS={x∈R:h(x)ﬀ0}.
12. First show thatf(0) = 0 andf(−x)=−f(x) for allx∈R; then note that
f(x−x
0)=f(x)−f(x 0). Consequentlyfis continuous at the pointx 0if
and only if it is continuous at 0. Thus, iffis continuous atx
0, then it is
continuous at 0, and hence everywhere.
13. First show thatf(0) = 0 and (by Induction) thatf(x)=cxforx∈N, and
hence also forx∈Z. Next show thatf(x)=cxforx∈Q. Finally, ifx/∈Q,
letx= lim(r
n) for some sequence inQ.
14. First show that eitherg(0) = 0 org(0) = 1. Next, ifg(α) = 0 for some
α∈Rand ifx∈R, lety:=x−αso thatx=α+y; henceg(x)=g(α+y)=
g(α)g(y) = 0. Thus, ifg(α) = 0 for someα, then it follows thatg(x)=0
for allx∈R.
Now suppose thatg(0) = 1 so thatg(c) = 0 for anyc∈R.Ifgis continuous
at 0, then givenε>0 there existsδ>0 such that if|h|<δ, then|g(h)−1|<
ε/|g(c)|. Sinceg(c+h)−g(c)=g(c)(g(h)−1), it follows that|g(c+h)−g(c)|=
|g(c)g(h)−1|<ε, provided|h|<δ. Thereforegis continuous atc.
15. Iff(x)≥g(x), then both expressions givenh(x)=f(x); and iff(x)≤g(x),
thenh(x)=g(x) in both cases.
Section 5.3
In this section, we establish some very important properties of continuous func-
tions. Unfortunately, students often regard these properties as being “obvious”,
so that one must convince them that if the hypotheses of the theorems are
dropped, then the conclusions may not hold. Thus, for example, if anyoneof the
three hypotheses [(i)Iis closed, (ii)Iis bounded, (iii)fis continuous at every
point ofI] of Theorem 5.3.2 is dropped, then the conclusion thatfis bounded
may not hold, even though the other two hypotheses are retained. Similarly for
Theorems 5.3.4 and 5.3.9. Thus, each theorem must be accompanied by examples.
In 5.3.7, we do not assume thatIis a closed bounded interval, but we work within
a closed bounded subinterval ofI.

36 Bartle and Sherbert
The proofs of Theorems 5.3.2 and 5.3.4 presented here are based on the
Bolzano-Weierstrass Theorem. In Section 5.5 diﬀerent proofs are presented based
on the concept of a “gauge”. In Chapter 11 these theorems are extended to general
“compact” sets inRby using the Heine-Borel Theorem.
Students often misunderstand Theorems 5.3.9 and 5.3.10, believing that the
image of an interval with endpointsf(a),f(b). Consequently, Figure 5.3.3 should
be stressed in an attempt to dispell this misconception. Also, examples can be
given to show that the continuous image of an interval (a, b) can be any type of
interval, and not necessarily an open interval or a bounded interval.
Sample Assignment: Exercises 1, 3, 5, 6, 7, 8, 10, 13, 15.
Partial Solutions:
1. Apply either the Boundedness Theorem 5.3.2 to 1/f, or the Maximum-
Minimum Theorem 5.3.4 to conclude that inff(I)>0.
Alternatively, ifx
n∈Isuch that 0<f(x n)<1/n, then there is a subse-
quence (x
nk) that converges to a pointx 0∈I. Sincef(x 0) = lim(f(x nk))=0,
we have a contradiction.
2. Iff(x
n)=g(x n) and lim(x n)=x 0, thenf(x 0) = lim(f(x n)) = lim(g(x n)) =
g(x
0).
3. Letx
1be arbitrary and letx 2∈Ibe such that|f(x 2)|≤
1
2
|f(x1)|. By Induc-
tion, choosex
n+1such that|f(x n+1)|≤
1
2
|f(xn)|≤

1
2

n
|f(x1)|. Apply the
Bolzano-Weierstrass Theorem to obtain a subsequence that converges to some
c∈I. Now show thatf(c)=0.
Alternatively, show that if the minimum value of|f|onIis not 0, then a
contradiction arises.
4. Suppose thatphas odd degreenand that the coeﬃcienta
nofx
n
is positive.
By 4.3.16, we have lim
x→∞
p(x)=∞and lim
x→−∞
p(x)=−∞. Hencep(α)<0 for
someα<0 andp(β)>0 for someβ>0. Therefore there is a zero ofpin
[α, β].
5. In the intervals [1.035, 1.040] and [−7.026,−7.025].
6. Note thatg(0) =f(0)−f(1/2) andg(1/2) =f(1/2)−f(1) =−g(0). Hence
there is a zero ofgat somec∈[0,1/2]. But if 0 =g(c)=f(c)−f(c+1/2),
then we havef(c)=f(c+1/2).
7. In the interval [0.7390, 0.7391].
8. In the interval [1.4687, 1.4765].
9. (a) 1, (b) 6.
10. 1/2
n
<10
−5
implies thatn>(5 ln 10)/ln 2≈16.61.Taken= 17.
11. Iff(w)<0, then it follows from Theorem 4.2.9 that there exists a
δ-neighborhoodV
δ(w) such thatf(x)<0 for allx∈V δ(w). But sincew<b,

Chapter 5 — Continuous Functions 37
this contradicts the fact thatw= supW. There is a similar contradiction if
we assume thatf(w)>0. Thereforef(w)=0.
12. Sincef(π/4)<1 whilef(0) = 1 andf(π/2)>1, it follows thatx
0∈(0,π/2).
If cosx
0>x
2
0
, then there exists aδ-neighborhoodV δ(x0)⊆Ion which
f(x) = cosx, so thatx
0is not an absolute minimum point forf.
13. Iff(x) = 0 for allx∈R, then all is trivial; hence, assume thatftakes on
some nonzero values. To be speciﬁc, supposef(c)>0 and letε:=
1
2
f(c),
and letM>0 be such that|f(x)|<εprovided|x|>M. By Theorem 5.3.4,
there existsc
∗
∈[−M,M] such thatf(c
∗
)≥f(x) for allx∈[−M,M] and we
deduce thatf(c
∗
)≥f(x) for allx∈R. To see that a minimum value need not
be attained, considerf(x):=1/(x
2
+ 1).
14. Apply Theorem 4.2.9 toβ−f(x).
15. If 0<a<b≤∞, thenf((a, b))=(a
2
,b
2
); if−∞ ≤a<b<0, thenf((a, b)) =
(b
2
,a
2
).Ifa<0<b, thenf((a, b)) is not an open interval, but equals [0,c)
wherec:= sup{a
2
,b
2
}. Images of closed intervals are treated similarly.
16. For example, ifa<0<bandc:= inf{1/(a
2
+1),1/(b
2
+1)}, theng((a, b)) =
(c,1]. If 0<a<b, theng((a, b))=(1/(b
2
+1),1/(a
2
+ 1)). Alsog([−1,1]) =
[1/2,1]. Ifa<b, thenh((a, b))=(a
3
,b
3
) andh((a, b])=(a
3
,b
3
].
17. Yes. Use the Density Theorem 2.4.8.
18. Iffis not bounded onI, then for eachn∈Nthere existsx
n∈Isuch that
|f(x
n)|≥n. Then a subsequence of (x n) converges tox 0∈I. The assumption
thatfis bounded on a neighborhood ofx
0leads to a contradiction.
19. Considerg(x):=1/xforx∈J:= (0,1).
Section 5.4
The idea of uniform continuity is a subtle one that often causes diﬃculties for stu-
dents. The point, of course, is that for a uniformly continuous functionf:A→R,
theδcan be chosen to dependonlyonεandnoton the points inA. The
Uniform Continuity Theorem 5.4.3 guarantees that every continuous function on
a closed bounded interval is uniformly continuous; however, a continuous func-
tion deﬁned on an interval may be uniformly continuous even when the interval
is not closed and bounded. For example, every Lipschitz function is uniformly
continuous, no matter what the nature of its domain is. A condition for a
function to be uniformly continuous on a bounded open interval is given in 5.4.8.
The extension of the Uniform Continuity Theorem to compact sets in given in
Chapter 11.
One interesting application of uniform continuity is the approximation of
continuous functions by “simpler” functions. Consequently we have included a
brief discussion of this topic here. The Weierstrass Approximation Theorem 5.4.14
is a fundamental result in this area and we have stated it without proof.

38 Bartle and Sherbert
Sample Assignment: Exercises 1, 2, 3, 6, 7, 8, 11, 12, 15.
Partial Solutions:
1. Since 1/x−1/u=(u−x)/xu, it follows that|1/x−1/u|≤(1/a
2
)|x−u|for
x, u∈[a,∞).
2. Ifx, u≥1, then|1/x
2
−1/u
2
|=(1/x
2
u+1/xu
2
)|x−u|≤2|x−u|,and it
follows thatfis uniformly continuous on [1,∞). Ifx
n:= 1/n, un:= 1/(n+ 1),
then|x
n−un|→0 but|f(x n)−f(u n)|=2n+1≥1 for alln,sofis not uni-
formly continuous on (0,∞).
3. (a) Letx
n:=n+1/n, u n:=n. Then|x n−un|→0, butf(x n)−f(u n)=
2+ 1/n
2
≥2 for alln.
(b) Letx
n:= 1/2nπ, u n:= 1/(2nπ+π/2). Note that|g(x n)−g(u n)|= 1 for
alln.
4. Show that|f(x)−f(u)|≤[(|x|+|u|)/(1+x
2
)(1+u
2
)]|x−u|≤(1/2+1/2)|x−u|=
|x−u|. (Note thatx →x/(1 +x
2
) attains a maximum of 1/2 atx= 1.)
5. Note that|(f(x)+g(x))−(f(u)+g(u))|≤|f(x)−f(u)|+|g(x)−g(u)|<ε
provided that|x−u|<inf{δ
f(ε/2),δ g(ε/2)}.
6. IfMis a bound for bothfandgonA, show that|f(x)g(x)−f(u)g(u)|≤
M|f(x)−f(u)|+M|g(x)−g(u)|for allx, u∈A.
7. Since lim
x→0
(sinx)/x= 1, there existsδ>0 such that sinx≥x/2 for 0≤x<δ.
Letx
n:= 2nπandu n:= 2nπ+1/n, so that sinx n= 0 and sinu n= sin(1/n).
Ifh(x):=xsinx, then|h(x
n)−h(u n)|=u nsin(1/n)≥(2nπ+1/n)/2n>π>0
for suﬃciently largen.
8. Givenε>0 there existsδ
f>0 such that|y−v|<δ fimplies|f(y)−f(v)|<ε.
Now chooseδ
g>0 so that|x−u|<δ gimplies|g(x)−g(u)|<δ f.
9. Note that|1/f(x)−1/f(u)|≤(1/k
2
)|f(x)−f(u)|.
10. There existsδ>0 such that if|x−u|<δ,x,u∈A, then|f(x)−f(u)|<1. If
Ais bounded, it is contained in the ﬁnite union of intervals of lengthδ.
11. If|g(x)−g(0)|≤K|x−0|for allx∈[0,1], then
√
x≤Kxforx∈[0,1]. But if
x
n:= 1/n
2
, thenKmust satisfyn≤Kfor alln∈N, which is impossible.
12. Givenε>0, choose 0<δ
1<1 so that|f(x)−f(u)|<εwhenever|x−u|<δ 1
andx, u∈[0,a+ 1]. Also choose 0<δ 2<1 so that|f(x)−f(u)|<εwhenever
|x−u|<δ
2andx, u∈[a,∞). Now letδ:= inf{δ 1,δ2}.If|x−u|<δ, then
sinceδ<1, eitherx, u∈[0,a+1] orx, u,∈[a,∞), so that|f(x)−f(u)|<ε.
13. Note that|f(x)−f(u)|≤|f(x)−g
ε(x)|+|g ε(x)−g ε(u)|+|g ε(u)−f(u)|.
14. Sincefis bounded on [0,p], it follows that it is bounded onR. Sincefis
continuous onJ:= [−1,p+ 1], it is uniformly continuous onJ. Now show
that this implies thatfis uniformly continuous onR.

Chapter 5 — Continuous Functions 39
15. Assume|f(x)−f(y)|≤K
f|x−y|and|g(x)−g(y)|≤K g|x−y|for allx, y
inA. (a)|(f(x)+g(x))−(f(y)+g(y))|≤|f(x)−f(y)|+|g(x)−g(y)|≤
(K
f+Kg)|x+y|.
(b) If|f(x)|≤B
fand|g(x)|≤B gfor allxinA, then
|f(x)g(x)−f(y)g(y)|=|f(x)g(x)−f(x)g(y)+f(x)g(y)−f(y)g(y)|
≤B
f|g(x)−g(y)|+B g|f(x)−f(y)|
≤(B
fKg+BgKf)|x−y|.
(c) Considerf(x)=x.
16. If|f(x)−f(y)|≤K|x−y|for allx, yinI, then has Lipschitz constantKonI.
Then for disjoint subintervals [x
k,yk],n=1,2,...,n, we have Σ|f(x k)−f(y k)|≤
ΣK|x
k−yk|, so that ifε>0 is given andδ=ε/nK, then Σ|f(x k)−f(y k)|≤ε.
Thusfis absolutely continuous onI.
Section 5.5
In this section we introduce the notion of a “gauge” which will be used in the
development of the generalized Riemann integral in Chapter 10. We will also use
gauges to give alternate proofs of the main theorems in Section 5.3 and 5.4, Dini’s
Theorem 8.2.6, and the Lebesgue Integrability Criterion in Appendix C.
Sample Assignment: Exercises 1, 2, 4, 6, 7, 9.
Partial Solutions:
1. (a) Theδ-intervals are [0−
1
4
,0+
1
4
]=[−
1
4
,
1
4
],[
1
2
−
1
4
,
1
2
+
1
4
]=[
1
4
,
3
4
] and
[
3
4
−
3
8
,
3
4
+
3
8
]=[
3
8
,
9
8
].
(b) The thirdδ-interval is [
3
10
,
9
10
] which does not contain [
1
2
,1].
2. (a) Yes. Sinceδ(t)≤δ
1(t), everyδ-ﬁne partition isδ 1-ﬁne.
(b) Yes. The thirdδ
1-interval is [
3
20
,
21
20
] which contains [
1
2
,1].
3. No. The ﬁrstδ
2-interval is [−
1
10
,
1
10
] and does not contain [0,
1
4
].
4. (b) Ift∈(
1
2
,1) then [t−δ(t),t+δ(t)]=[−
1
2
+
3
2
t,
1
2
+
1
2
t]⊂(
1
4
,1).
5. Routine veriﬁcation.
6. We could have two subintervals havingcas a tag with one of them not
contained in theδ-interval aroundc. Consider constant gaugesδ
ﬃ
:= 1 on
[0, 1] andδ
ﬃﬃ
:=
1
2
on [1, 2], so thatδ(1) =
1
2
.If
˙
P
ﬃ
consists of the single pair
([0,1],1), it isδ
ﬃ
-ﬁne. However,
˙
P
ﬃ
is notδ-ﬁne.
7. Clearlyδ
∗
(t)>0 so thatδ
∗
is a gauge on [a, b]. If
˙
P:={([a, x
1],t1), ...([x k−1,c],tk),([c, xk+1],tk+1), ...,([x n,b],tn)}

40 Bartle and Sherbert
isδ
∗
-ﬁne, then it is clear that
˙
P
ﬃ
:={([a, x 1],t1), ...,([x k−1,c],tk)}is a
δ
ﬃ
-ﬁne partition of [a, c] and
˙
P
ﬃﬃ
:={([c, x k+1],tk+1), ...,([x n,b],tn)}is a
δ
ﬃﬃ
-ﬁne partition of [c, b]. Evidently
˙
P=
˙
P
ﬃ
∪
˙
P
ﬃﬃ
.
8. (a) If [a, b]aδ-ﬁne partition
˙
P
ﬃ
and [c, b] has aδ-ﬁne partition
˙
P
ﬃﬃ
, then
˙
P
ﬃ
∪
˙
P
ﬃﬃ
is aδ-ﬁne partition of [a, b], contrary to hypothesis.
(b) LetI
1be [a, c]ifitdoesnothaveaδ-ﬁne partition; otherwise, letI 1be
[c, b], so the length ofI
1is (b−a)/2. Now bisectI 1and letI 2, which has
length (b−a)/2
2
, be an interval that has noδ-ﬁne partition. Continue this
process by Induction.
(c) By the Nested Intervals Theorem there exists a common pointξ. By the
Archimedean Property there existsp∈Nsuch that (b−a)/2
p
<δ(ξ). Since
ξ∈I
pand the length ofI pis (b−a)/2
p
, it follows thatI p⊂[ξ−δ(ξ),ξ+δ(ξ)].
9. The hypothesis thatfis locally bounded presents us with a gaugeδ.If
{([x
i−1,xi],ti)}
n
i=1
is aδ-ﬁne partition of [a, b] andM iis a bound for|f|on
[x
i−1,xi], letM:= sup{M i:i=1, ...,n}.
10. The hypothesis thatfis locally increasing presents us with a gaugeδ.If
{([x
i−1,xi],ti)}
n
i=1
is aδ-ﬁne partition of [a, b], thenfis increasing on each
interval [x
i−1,xi]. By Induction it follows thatf(x i)≤f(x j) fori<j.Ifx<y
belong to [a, b], thenx∈[x
i−1,xi] andy∈[x j−1,xj] wherei≤j.Ifi=j, the
fact thatfis increasing on [x
i−1,xi] implies thatf(x)≤f(y). Ifi<j, then
f(x)≤f(x
i)≤f(x j−1)≤f(y).
Section 5.6
The collection of monotone functions is a special, but very useful class of functions.
This is particularly the case since most functions that arise in elementary analysis
are either monotone, or their domains can be written as a union of intervals
on which their restrictions are monotone. Theorem 5.6.4 shows that a monotone
function is automatically continuous except (at most) at a countable set of points.
It will also be seen in Theorem 5.6.5 that continuous strictly monotone
functions have continuous strictly monotone inverse functions.
Sample Assignment: Exercises 1, 2, 4, 5, 7, 10, 12.
Partial Solutions:
1. Ifx∈[a, b], thenf(a)≤f(x).
2. Ifx
1≤x2, thenf(x 1)≤f(x 2) andg(x 1)≤g(x 2), whencef(x 1)+g(x 1)≤
f(x
2)+g(x 2).
3. Note that (fg)(0) = 0>(fg)(1/2) =−1/4.
4. If 0≤f(x
1)≤f(x 2) and 0≤g(x 1)≤g(x 2), thenf(x 1)g(x1)≤f(x 2)g(x1)≤
f(x
2)g(x2).

Chapter 5 — Continuous Functions 41
5. IfL:= inf{f(x):x∈(a, b]}andε>0, then there existsx
ε∈(a, b] withL≤
f(x
ε)<L+ε. Sincefis increasing, thenL≤f(x)<L+εforx∈(a, x ε];
hence lim
x→a+
fexists and equalsL.
Conversely, ifK:= lim
x→a+
f, then givenε>0, there existsδ>0 such
that ifx∈(a, a+δ), thenK−ε<f(x)<K+ε. It follows from this that
K−ε≤L< K+ε; sinceε>0 is arbitrary, we haveK=L.
6. Iffis continuous atc, then lim(f(x
n)) =f(c), sincec= lim(x n). Conversely,
since 0≤j
f(c)≤f(x 2n)−f(x 2n+1), it follows thatj f(c)=0, sofis continu-
ous atc.
7. It follows from Exercises 2.4.4, 2.4.6 and the Principle of the Iterated Inﬁma,
(analogous to the result in Exercise 2.4.12), that
j
f(c) = inf{f(y):y∈I,c<y}−sup{f(x):x∈I,x<c}
= inf{f(y):y∈I,c<y}+ inf{−f(x):x∈I,x<c}
= inf{f(y)−f(x):x, y∈I,x<c<y}
8. Letx
1∈Ibe such thaty=f(x 1) andx 2∈Ibe such thaty=g(x 2). Ifx 2≤x1,
theny=g(y
2)<f(x 2)≤f(x 1)=y, a contradiction.
9. Ifx∈Iis rational, thenf(x)=xis also rational sof(f(x)) =f(x)=x;if
y∈Iis irrational, thenf(y)=1−yis irrational sof(f(y)) =f(1−y)=1−
(1−y)=y. Suppose thatx
1 =x2,xj∈I;ifx 1∈Qandx 2/∈Q, thenf(x 1)∈Q
andf(x
2)/∈Q, which implies thatf(x 1) =f(x 2). The other cases are similar.
Since|f(x)−1/2|=|x−1/2|,thenfis continuous at 1/2. Ifx =1/2,x∈Q,
take a sequence (y
n) of irrationals converging tox, so thatf(y n)=
1−y
n→1−x =x. Similarly for the casex =1/2,x /∈Q.
10. Iffhas an absolute maximum atc∈(a, b), and iffis injective, we have
f(a)<f(c) andf(b)<f(c). Eitherf(a)≤f(b)orf(b)<f(a). In the ﬁrst
case, eitherf(a)=f(b)orf(a)<f(b)<f(c), whence there existsb
ﬃ
∈(a, c)
such thatf(b
ﬃ
)=f(b). Either possibility contradicts the assumption that
fis injective. The casef(b)<f(a) is similar.
11. Note thatf
−1
is continuous at every point of its domain [0,1]∪(2,3]. The
functionfis not continuous atx=1.
12. Leta∈(0,1) be arbitrary. Iff(a)<f(0), then there existsa
ﬃ
∈(a,1) with
f(a
ﬃ
)=f(0), a contradiction. Alsof(a)=f(0) is excluded by hypothesis.
Therefore we must havef(0)<f(a),and a similar argument yieldsf(a)<f(1).
Ifb∈(a,1) is given, thenf(b)<f(a) implies that there existsa
ﬃﬃ
∈(b,1) with
f(a)=f(a
ﬃﬃ
), a contradiction. Sincef(b)=f(a) is excluded, we must have
f(b)>f(a).
13. Assume thathis continuous on [0,1] and letc
1<c2be the two points in [0,1]
wherehattains its supremum. If 0<c
1, choosea 1,a2such that 0<a 1<c1<
a
2<c2. Letksatisfy sup{h(a 1),h(a 2)}<k<h(c 1)=h(c 2); then there exist

42 Bartle and Sherbert
three numbersb
jsuch thata 1<b1<c1<b2<a2<b3<c2wherek=h(b j),
a contradiction. Now consider the points wherehattains its inﬁmum.
14. Letx>0 and consider the casem, p, n, q∈N. Lety:=x
1/n
andz:=x
1/q
so
thaty
n
=x=z
q
, whence (by Exercise 2.1.26)y
np
=x
p
=z
qp
. Sincenp=mq,
we have (y
m
)
q
=y
mq
=z
pq
=(z
p
)
q
, from which it follows thaty
m
=z
p
,or
(x
1/n
)
m
=(x
1/q
)
p
,orx
m/n
=x
p/q
. Now consider the case wherem, p∈Z.
15. Letx>0 and consider the case wherer=m/nands=p/q, wherem, n, p,
q∈N. Sincer=mq/nqands=pn/qn, it follows from the preceding exercise
thatx
r
=(x
1/nq
)
mq
andx
s
=(x
1/nq
)
pn
so that (by Exercise 2.1.26)x
r
x
s
=
(x
1/nq
)
mq+pn
=x
(mq+pn)/nq
=x
r+s
. Similarly,x
r
=(x
1/n
)
m
>0 and ify>0,
then (by 5.6.7)y
s
=(y
p
)
1/q
so that (x
r
)
s
= (((x
1/n
)
m
)
p
)
1/q
. This implies
that ((x
r
)
s
)
q
=(x
1/n
)
mp
=(x
mp
)
1/n
so that ((x
r
)
s
)
qn
=x
mp
, whence (x
r
)
s
=
x
mp/qn
=x
rs
. Now consider the case wherem, p∈Z.

CHAPTER 6
DIFFERENTIATION
The basic properties and applications of the derivative are given in the ﬁrst two
sections of this chapter. Section 6.1 is a survey of the techniques of diﬀerentiation
from a rigorous viewpoint. Since the students will be familiar with most of the
results (though not the proofs), the section can be covered reasonably quickly.
Section 6.2 contains material that is new to students, since in introductory calculus
courses the Mean Value Theorem is not usually given the emphasis it deserves.
Sections 6.3 and 6.4 are optional and can be discussed in either order and to
whatever depth that time permits.
Section 6.1
This section contains the calculational rules of diﬀerentiation that students learn
and use in introductory calculus courses. However, the emphasis here is on the
rigorous establishment of these results rather than on the development of calcu-
lational skills.
The topic that students will ﬁnd troublesome is the diﬀerentiation of com-
posite and inverse functions. We feel that the use of Carath´eodory’s Theorem
6.1.5 is a considerable simpliﬁcation of the proofs of these results.
Sample Assignment: Exercises 1(a,b), 2, 4, 5, 9, 11, 13, 15.
Partial Solutions:
1. (a)f
ﬃ
(x) = lim
h→0
[(x+h)
3
−x
3
]/h= lim
h→0
(3x
2
+3xh+h
2
)=3x
2
,
(b)g
ﬃ
(x) = lim
h→0
1
h
π
1
x+h
−
1
x
α
= limh→0
−1
x(x+h)
=
−1
x
2
,
(c)h
ﬃ
(x) = lim
h→0
√
x+h−
√
x
h
= lim h→0
1
√
x+h+
√
x
=
1
2
√
x
,
(d)k
ﬃ
(x) = lim
h→0
1/
√
x+h−1/
√
x
h
= lim h→0
−1
√
x+h
√
x(
√
x+h+
√
x)
=
−1
2x
√
x
.
2. lim
x→0
(f(x)−f(0))/(x−0) = lim
x→0
x
1/3
/x= lim
x→0
1/x
2/3
does not exist.
3. (a) (αf)
ﬃ
(c) = lim
x→c
αf(x)−αf(c)
x−c
=αlim x→c
f(x)−f(c)
x−c
=αf
ﬃ
(c),
(b) (f+g)
ﬃ
(c) = lim
x→c
(f(x)+g(x))−(f(c)+g(c))
x−c
= lim
x→c
f(x)−f(c)
x−c
+ lim x→c
g(x)−g(c)
x−c
=f
ﬃ
(c)+g
ﬃ
(c).
43

44 Bartle and Sherbert
4. Note that|f(x)/x|≤|x|forx∈R.
5. (a)f
ﬃ
(x)=(1−x
2
)/(1 +x
2
)
2
,
(b)g
ﬃ
(x)=(x−1)/
√
5−2x+x
2
,
(c)h
ﬃ
(x)=m(sinx
k
)
m−1
(cosx
k
)(kx
k−1
),
(d)k
ﬃ
(x)=2xsec
2
(x
2
).
6. The functionf
ﬃ
is continuous forn≥2 and is diﬀerentiable forn≥3.
7. By deﬁnitiong
ﬃ
(c) = lim
h→0
|f(c+h)|/h, if this limit exists. If 0 =|f
ﬃ
(c)|=
lim
h→0
|f(c+h)/h|, it follows thatg
ﬃ
(c)=0. Iff
ﬃ
(c)=L = 0, then we have
lim(f(c±1/n)/(±1/n)) =L, while lim(|f(c±1/n)|/(±1/n))] =±L, so that
|f|
ﬃ
(c) does not exist.
8. (a)f
ﬃ
(x)=2 forx>0;f
ﬃ
(x)=0 for−1<x<0; andf
ﬃ
(x)=−2 forx<−1,
(b)g
ﬃ
(x)=3 ifx>0;g
ﬃ
(x)=1 ifx<0;g
ﬃ
(0) does not exist,
(c)h
ﬃ
(x)=2|x|for allx∈R,
(d)k
ﬃ
(x)=(−1)
n
cosxfornπ<x<(n+1)π,n∈Z;k
ﬃ
(nπ) does not exist,
(e)p
ﬃ
(0)=0; ifx = 0, thenp
ﬃ
(x) does not exist.
9. Iffis an even function, thenf
ﬃ
(−x) = lim
h→0
[f(−x+h)−f(−x)]/h=
−lim
h→0
[f(x−h)−f(x)]/(−h)=−f
ﬃ
(x).
10. Ifx = 0, theng
ﬃ
(x)=2xsin(1/x
2
)−(2/x) cos(1/x
2
). Moreover,g
ﬃ
(0) =
lim
h→0
hsin(1/h
2
)=0. Ifx n:= 1/
√
2nπ, thenx n→0 and|g
ﬃ
(xn)|=2
√
2nπ,
sog
ﬃ
is unbounded in every neighborhood of 0.
11. (a)f
ﬃ
(x)=2/(2x+ 3), (b)g
ﬃ
(x)=6(L(x
2
))
2
/x,
(c)h
ﬃ
(x)=1/x, (d) k
ﬃ
(x)=1/(xL(x)).
12.r>1.
13. Many examples are possible. For example, letf(x):=xforxrational and
f(x) := 0 forxirrational.
14. 1/h
ﬃ
(0)=1/2,1/h
ﬃ
(1)=1/5 and 1/h
ﬃ
(−1)=1/5.
15.D[Arccosy]=1/D[cosx]=−1/sinx=−1/
Σ
1−y
2
.
16.D[Arctany]=1/D[tanx]=1/sec
2
x=1/(1 +y
2
).
17. Givenε>0, letδ(ε)>0 be such that if 0<|w−c|<δ(ε),w∈I, then
|f(w)−f(c)−(w−c)f
ﬃ
(c)|<ε|w−c|. Now takew=uandw=vas described
and subtract and add the termf(c)−f
ﬃ
(c)cand use the Triangle Inequality
to get
|f(v)−f(u)−f
ﬃ
(c)(v−u)|≤|f(v)−f(c)−f
ﬃ
(c)(v−c)|
+|f(c)−f(u)−f
ﬃ
(c)(c−u)|≤ε|v−c|+ε|c−u|.

Chapter 6 — Differentiation 45
Sincev−c≥0 andc−u≥0, then|v−c|=v−cand|c−u|=c−u, so that
the ﬁnal term equalsε(v−c+c−u)=ε(v−u).
Section 6.2
The Mean Value Theorem is stated for a functionfon an interval [a, b]. However,
many of its applications use intervals of the form [a, x]or[x
1,x2] wherexorx 1,
x
2are points in [a, b]. The shift from a “ﬁxed interval” to what seems to be a
“variable interval” can cause confusion for some students. A word of explanation
when this ﬁrst occurs will help to alleviate this confusion.
WARNING: Exercises 16 and 18 are rather diﬃcult.
Sample Assignment: Exercises 2(a, b), 3(a, b), 6, 7, 9, 10, 12, 13, 17.
Partial Solutions:
1. (a) Increasing on [3/2,∞), decreasing on (−∞,3/2],
(b) Increasing on (−∞,3/8], decreasing on [3/8,∞),
(c) Increasing on (−∞,−1] and [1,∞),
(d) Increasing on [0,∞).
2. (a)f
ﬃ
(x)=1−1/x
2
. Relative minimum atx= 1; relative maximum atx=−1,
(b)g
ﬃ
(x)=(1+x)(1−x)/(1 +x
2
)
2
. Relative minimum atx=−1; relative
maximum atx=1,
(c)h
ﬃ
(x)=1/2
√
x−1/
√
x+ 2. Relative maximum atx=2/3,
(d)k
ﬃ
(x)=2(x
3
−1)/x
3
. Relative minimum atx=1.
3. (a) Relative minima atx=±1; relative maxima atx=0,±4,
(b) Relative maximum atx= 1; relative minima atx=0,2,
(c) Relative minima atx=−2,3; relative maximum atx=2,
(d)k
ﬃ
(x)=4(x−6)/3(x−8)
2/3
. Relative minimum atx= 6; relative maxima
atx=0,9.
4.x=(1/n)(a
1+···+a n).
5. Show thatf
ﬃ
(x)<0 forx>1. Thenfis strictly decreasing on [1,∞) so that
f(a/b)<f(1) fora>b>0.
6. Ifx<y, there existscin (x, y) such that|sinx−siny|=|cosc||y−x|.
7. There existscwith 1<c<xsuch that lnx=(x−1)/c. Now use the inequal-
ity 1/x <1/c <1.
8. Ifh>0 anda+h<b, there existsc
h∈(a, a+h) such thatf(a+h)−f(a)=
hf
ﬃ
(ch). Sincec h→aash→0 + , it follows thatf
ﬃ
(a) = lim
h→0+
[f(a+h)−
f(a)]/h= lim
h→0+
f
ﬃ
(ch)=A. Now considerh<0.
9.f(x)=x
4
(2 + sin(1/x))>0 for allx =0, sofhas an absolute minimum at
x= 0. We havef
ﬃ
(x)=8x
3
+4x
3
sin(1/x)−x
2
cos(1/x) forx = 0. Now verify
thatf
ﬃ
(1/2nπ)<0 forn≥2 andf
ﬃ
(2/(4n+1)π)>0 forn≥1.

46 Bartle and Sherbert
10.g
ﬃ
(0) = lim
x→0
(1+2xsin(1/x))=1+0=1,and ifx = 0, theng
ﬃ
(x)=1+
4xsin(1/x)−2 cos(1/x). Now show thatg
ﬃ
(1/2nπ)<0 and that we have
g
ﬃ
(2/(4n+1)π)>0 forn∈N.
11. For example,f(x):=
√
x.
12. Apply Darboux’s Theorem 6.2.12. Ifg(x):=aforx<0,g(x):=x+bfor
x≥0, wherea, bare any constants, theng
ﬃ
(x)=h(x) forx =0.
13. Ifx
1<x2, then there existsc∈(x 1,x2) such thatf(x 2)−f(x 1)=
(x
2−x1)f
ﬃ
(c)>0.
14. Apply Darboux’s Theorem 6.2.12.
15. Suppose that|f
ﬃ
(x)|≤Kforx∈I.Forx, y∈I, apply the Mean Value
Theorem to get|f(x)−f(y)|=|(x−y)f
ﬃ
(c)|≤K|x−y|.
16. (a) Givenε>0 there existsn
ε∈Nsuch that ifx≥n ε, then|f
ﬃ
(x)−b|<ε.
Hence ifx≥n
εandh>0, there existsy x∈(x, x+h) such that

f(x+h)−f(x)
h
−b

=|f
ﬃ
(yx)−b|<ε.
Sinceε>0 is arbitrary, then lim
x→∞
(f(x+h)−f(x))/h=b.
(b) Assume thatb = 0 and letε<|b|/2.Letn
εbe as in part (a). Since lim
x→∞
f
exists, we may also assume that ifx, y≥n
ε, then|f(x)−f(y)|<ε. Hence
there existsx
εin (n ε,nε+ 1) such thatε>|f(n ε+1)−f(n ε)|=|f
ﬃ
(xε)|≥
|b|/2. Sinceε>0 is arbitrary, the hypothesis thatb = 0 is contradicted.
(c) Ifx≥n
ε, then there existsy ε∈(nε,x) such thatf(x)−f(n ε)=
(x−n
ε)f
ﬃ
(yε), so that we have
f(x)/x−b=f
ﬃ
(yε)−b+f(n ε)/x−n εf
ﬃ
(yε)/x.
Sincey
ε>nε, we have|f
ﬃ
(yε)−b|<ε. Moreover|f(n ε)/x|<εifxis suﬃ-
ciently large; sincef
ﬃ
is bounded on [n ε,∞), then|n εf
ﬃ
(yε)/x|<εifxis
suﬃciently large. Therefore, lim
x→∞
f(x)/x=b.
17. Apply the Mean Value Theorem to the functiong−fon [0,x].
18. Givenε>0,letδ=δ(ε) be as in Deﬁnition 6.1.1, and letx<c<ybe such
that 0<|x−y|<δ. Sincef(x)−f(y)=f(x)−f(c)+f(c)−f(y), a simple
calculation shows that
f(x)−f(y)
x−y
=
x−c
x−y
·
f(x)−f(c)
x−c
+
c−y
x−y
·
f(y)−f(c)
y−c
.

Chapter 6 — Differentiation 47
Since both (x−c)/(x−y) and (c−y)/(x−y) are positive and have sum 1,
it follows that

f(x)−f(y)
x−y
−f
ﬃ
(c)

≤
π
x−c
x−y
α

f(x)−f(c)
x−c
−f
ﬃ
(c)

+
π
c−y
x−y
α

f(y)−f(c)
y−c
−f
ﬃ
(c)

<
π
x−c
x−y
+
c−y
x−y
α
ε=ε.
Note that if one (but not both) ofxandyequalc, the conclusion still holds.
19. Letx, y∈I,x =y; then
f
ﬃ
(x)−f
ﬃ
(y)=f
ﬃ
(x)−
f(x)−f(y)
x−y
+
f(x)−f(y)
x−y
−f
ﬃ
(y),
so that

f
ﬃ
(x)−f
ﬃ
(y)

≤

f
ﬃ
(x)−
f(x)−f(y)
x−y

+

f(x)−f(y)
x−y
−f
ﬃ
(y)

.
Iffis uniformly diﬀerentiable onI, givenε>0 there existsδ>0 such that
if 0<|x−y|<δ,x,y∈I, then both terms on the right side are less thanε.
Hence we have|f
ﬃ
(x)−f
ﬃ
(y)|<2εfor|x−y|<δ, x,y∈I, whence we con-
clude thatf
ﬃ
is (uniformly) continuous onI.
20. (a,b) Apply the Mean Value Theorem.
(c) Apply Darboux’s Theorem to the results of (a) and (b).
Section 6.3
The proofs of the various cases of L’Hospital’s Rules range from fairly trivial to
rather complicated. The only really diﬃcult argument in this section is the proof
of Theorem 6.3.5, which deals with the case∞/∞. This requires a more subtle
analysis than the other cases.
This section may be regarded as optional. Students are already familiar with
the mechanics of L’Hospital’s Rules.
Sample Assignment: Exercises 1, 2, 4, 6, 7(a,b), 8(a,b), 9(a,b), 13, 14.
Partial Solutions:
1.A=B(lim
x→c
f(x)/g(x))=0.
2. IfA>0, thenfis positive on a neighborhood ofcand lim
x→c
(g(x)/f(x))=0.
Sinceg(x)/f(x)>0 on a neighborhood ofc, we use the fact thatf(x)/g(x)=
1/[g(x)/f(x)] to get a limit of∞.

48 Bartle and Sherbert
3. In fact, ifx∈(0,1] thenf(x)/g(x) = sin(1/x), which does not have a limit
at 0. [Note that 6.3.1 cannot applied be applied sinceg
ﬃ
(0) = 0, and 6.3.3
cannot be applied sincef
ﬃ
(x)/g
ﬃ
(x) does not have a limit at 0.]
4. Note thatf
ﬃ
(0) = 0, but thatf
ﬃ
(x) does not exist ifx =0.
5. Recall that lim
x→0
(sinx)/x=1,but that lim
x→0
cos(1/x) does not exist.
6. (a) lim
x→0
e
x
−e
−x
−2
1−cosx
= lim x→0
e
x
+e
−x
sinx
= lim x→0
e
x
+e
−x
cosx
=2.
(b) lim
x→0
x
2
−sin
2
x
x
4
= lim
x→0
2x−2 sinxcosx
4x
3
= lim
x→0
2x−sin 2x
4x
3
= lim
x→0
1−cos 2x
6x
2
= lim
x→0
2 sin 2x
12x
= limx→0
4 cos 2x
12
=
1
3
.
7. (a) 1, (b) 1, (c) 0, (d) 1/3.
8. (a) 1, (b)∞, (c) 0, (d) 0.
9. (a) 0, (b) 0, (c) 0, (d) 0.
10. (a) 1, (b) 1, (c) e
3
, (d) 0.
11. (a) 1, (b) 1, (c) 1, (d) 0.
12. Leth(x):=e
x
f(x).Thenh
ﬃ
(x)=e
x
(f(x)+f
ﬃ
(x)),so that lim
x→∞
h
ﬃ
(x)/e
x
=
lim
x→∞
(f(x)+f
ﬃ
(x)) =L, by hypothesis. Giveε>0, there existsα>0 such
thatL−ε/2<h
ﬃ
(x)/e
x
<L+ε/2 for allx>α.Ifα<y<x, then by 6.3.2
there existsc>αwith
h(x)−h(y)
e
x
−e
y
=
h
ﬃ
(c)
e
c
,
and thereforeL−ε/2<(h(x)−h(y))/(e
x
−e
y
)<L+ε/2. But sincee
x
−e
y
>0,
this implies that
e
x
−e
y
e
x
·(L−ε/2)<
h(x)−h(y)
e
x
<
e
x
−e
y
e
x
·(L+ε/2).
Addh(y)/e
x
to all sides and rearrange terms to get
(L−ε/2) +
h(y)−e
y
(L−ε/2)
e
x
<
h(x)
e
x
<(L+ε/2) +
h(y)−e
y
(L+ε/2)
e
x
.
For ﬁxedy, we note that lim
x→∞
[h(y)−e
y
(L±ε/2)]/e
x
= 0. Sinceh(x)/e
x
=
f(x), it follows that for suﬃciently largexwe haveL−ε<f(x)<L+ε.
Therefore lim
x→∞
f(x)=L, which implies that lim
x→∞
f
ﬃ
(x) = lim
x→∞
(f(x)+f
ﬃ
(x))−
lim
x→∞
f(x)=L−L=0.
[Note. Ife
x
is replaced by a functiong(x) such thatg
ﬃ
(x)>0 for large
values ofx, then the above argument can be modiﬁed slightly to prove the

Chapter 6 — Differentiation 49
following version of L’Hospital’s Rule: Ifhandgare diﬀerentiable func-
tions on (0,∞) that satisfy lim
x→∞
h
ﬃ
(x)/g
ﬃ
(x)=Land lim
x→∞
g
ﬃ
(x)=∞,then
lim
x→∞
h(x)=L.]
13. The limit is 1.
14. lim
x→c
x
c
−c
x
x
x
−c
c
= lim
x→c
cx
c−1
−(lnc)c
x
(1 + lnx)x
x
=
1−lnc
1+ lnc
.
Section 6.4
The applications of Taylor’s Theorem are similar in spirit to those of the Mean
Value Theorem, but the technical details can be more complicated since higher
order derivatives are involved. Instead of estimatingf
ﬃ
, the use of Taylor’s
Theorem usually requires the estimation of the remainderR
n.
The applications that are presented here are independent of one another and
they need not all be covered to illustrate the use and importance of Taylor’s
Theorem. If Newton’s Method is discussed, students should be encouraged to
program the algorithm on a computer or programmable calculator; comparison of
the rate of convergence with the bisection method of locating roots is instructive.
Sample Assignment: Exercises 1, 2, 4, 5, 7, 8, 12, 14(a,b), 19, 20, 23.
Partial Solutions:
1.f
(2n−1)
(x)=(−1)
n
a
2n−1
sinaxandf
(2n)
(x)=(−1)
n
a
2n
cosaxforn∈N.
2.g
ﬃ
(x)=3x
2
forx≥0,g
ﬃ
(x)=−3x
2
forx<0, andg
ﬃﬃ
(x)=6|x|forx∈R.
3. Use the relation:

n+1
k

=

n
k

+

n
k−1

for 0≤k≤n, wherek,n∈N.
4. Apply Taylor’s Theorem tof(x):=
√
1+xatx 0:= 0 and note that
R
1(x)<0 andR 2(x)>0 forx>0.
5. 1.095<
√
1.2<1.1 and 1.375<
√
2<1.5.
6.R
2(0.2)<0.0005 andR 2(1)<0.0625.
7.R
2(x)=(1/6)(10/27)(1 +c)
−8/3
x
3
<(5/81)x
3
, where 0<c<x.
8.R
n(x)=e
c
(x−x 0)
n+1
/(n+ 1)!→0asn→∞.
9.|R
n(x)|≤|x−x 0|
n
/n!→0ann→∞.
10. Use Induction to showh
(n)
(0) = 0 forn∈N.Ifx = 0, thenh
(n)
(x)isthe
sum of terms of the forme
−1/x
2
/x
k
; therefore, ifh
(n)
(0) = 0, thenh
(n+1)
(0) =
lim
x→0
h
(n)
(x)/x= 0. SinceP n(x) = 0 for allxand alln, whileh(x) = 0 for
x = 0, the remainderR
n(x) cannot converge to 0 forx =0.
11. Withn=4, ln1.5=0.40; withn=7, ln1.5=0.405.
12. UseP
6(x) and note that 7! = 5040.
13. Forf(x)=e
x
atx0= 0, the remainder atx= 1 satisﬁes the inequalityR n(1)≤
3/(n+ 1)!<10
−7
ifn≥10.P 10(1)=2.718 281 8 to seven places.

50 Bartle and Sherbert
14. (a) No. (b) No. (c) No. (d) Relative minimum.
15. Apply the Mean Value Theorem tofno [a, x
0] and on [b, x 0] getc 1and
c
2such thatf
ﬃ
(c1)=f
ﬃ
(c2). Now apply the Mean Value Theorem tof
ﬃ
on
[c
1,c2].
16. To obtain the formula, apply 6.3.3 and 6.3.1.
17. Apply Taylor’s Theorem tofatx
0=cto getf(x)≥f(c)+f
ﬃ
(c)(x−c).
18. Apply Taylor’s Theorem tofand then togatx
0=c. Then form the quotient
and use the continuity of thenth derivatives.
19. Sincef(2)<0 andf(2.2)>0, there is a zero offin [2.0, 2.2]. The value of
x
4is approximately 2.094 551 5.
20.r
1≈1.452 626 88 andr 2≈−1.164 035 14.
21.r≈1.324 717 96.
22.r
1≈0.158 594 34 andr 2≈3.146 193 22.
23.r
1≈0.5 andr 2≈0.809 016 99.
24.r≈0.739 085 13.

CHAPTER 7
THE RIEMANN INTEGRAL
Students will, of course, have met with the Riemann integral in calculus, although
few of them would be able to deﬁne it with any precision. The approach used here
is almost certain to be the same as that used in their calculus course, although it
will probably come as news to the students that the subintervals in the partitions
do not need to have equal length.
If the students have a strong background, it is possible to go quickly through
this chapter and then discuss part of Chapter 10, dealing with the generalized
Riemann integral. However, for most classes in a one semester course, all of the
time available may be needed to cover this chapter. In that case, Chapter 10 can
be assigned as a “extra topic” to special students.
Since the most important results in this chapter are the Fundamental Theo-
rems given in Section 7.3, discussion should be focussed to lead to these results.
Section 7.4 is an optional section on the Darboux approach to the integral
using upper and lower integrals. The relative merits of the two approaches are
discussed in the introduction to the section. Section 7.5 deals with methods of
approximating integrals.
Section 7.1
Most of the results will be familiar to students. Discuss the examples carefully and
sample some of the proofs. Leave time for a discussion of some of the exercises.
Sample Assignment: Exercises 1(a,c), 2(a,c), 6, 9, 10, 12, 14.
Partial Solutions:
1. (a)P
1= 2, (b)P 2= 2 (c)P 3=1.4, (d)P 4=2.
2. (a) 0
2
·1+1
2
·1+2
2
·2=0+1+8=9,
(b) 1
2
·1+2
2
·1+4
2
·2=1+4+32=37,
(c) 0
2
·2+2
2
·1+3
2
·1=0+4+9=13,
(d) 2
2
·2+3
2
·1+4
2
·1=8+9+16=33.
3. Deﬁnition 7.1.1 requires that if
˙
P<δ
ε, then|S(f;
˙
P)−L|<ε. Therefore, if

˙
P≤δ
ε/2, then
˙
P<δ εso that|S(f;
˙
P)−L|≤ε. Hence we takeδ
ﬃ
ε
:=δε/2.
On the other hand, if
˙
P≤η
εimplies that|S(f;
˙
P)−L|≤ε, we setδ ε:=
(1/2)η
ε/2. Then if
˙
P≤δ εthen
˙
P<η
ε/2so that|S(f;
˙
P)−L|≤ε/2<ε.
4. (b) Ifu∈U
2, thenu∈[x i−1,xi] with tagt i∈[1,2], so that (i)x i−1≤ti≤2
which implies thatu≤x
i≤xi−1+
˙
P≤2+
˙
Pand (ii) 1≤t i≤xiwhich
implies that 1?
˙
P≤x
i?
˙
P≤x i−1≤u. Thereforeubelongs to [1?
˙
P,
2+
˙
P].
51

52 Bartle and Sherbert
On the other hand, if 1 +
˙
P ≤v≤2?
˙
Pandv∈[x
i−1,xi], then
(i) 1 +
˙
P ≤x
iwhich implies that 1≤x i?
˙
P ≤x i−1≤tiand (ii)x i−1≤
2?
˙
Pwhich implies thatt
i≤xi≤xi−1+
˙
P ≤2. Therefore we get
t
i∈[1,2].
5. (a) Ifu∈[x
i−1,xi], thenx i−1≤uso thatc 1≤ti≤xi≤xi−1+
˙
Pwhence
c
1?
˙
P≤x i−1≤u. Alsou≤x iso thatx i?
˙
P≤x i−1≤ti≤c2, whence
u≤x
i≤c2+
˙
P.
(b) Ifc
1+
˙
P≤v≤x ithenc 1≤xi?
˙
P≤x i−1and ifv≤c 2?
˙
P, then
x
i≤xi−1+
˙
P≤c 2. Thereforec 1≤xi−1≤ti≤xi≤c2.
6. (a) If
˙
Pis a tagged partition of [0, 2], let
˙
P
1be the subset of
˙
Phaving tags
in [0, 1], and let
˙
P
2be the subset of
˙
Phaving tags in [1, 2]. The union of
the subintervals in
˙
P
1contains the interval [0,1?
˙
P] and is contained in
[0,1+
˙
P], so that 2(1?
˙
P)≤S(f;
˙
P
1)≤2(1 +
˙
P). Similarly, the union
of the subintervals in
˙
P
2contains [1 +
˙
P,2] and is contained in [1?
˙
P,2],
so that 1?
˙
P ≤S(f;
˙
P
2)≤1+
˙
P. Therefore 3−3
˙
P ≤S(f;
˙
P)=
S(f;
˙
P
1)+S(f;
˙
P 2)≤3+3
˙
P, whence|S(f;
˙
P)−3|≤3
˙
P, and we should
take
˙
P<ε/3.
(b) If
˙
P
0is the subset of
˙
Phaving tags at 1, then the union of the (at
most two) subintervals in
˙
P
0is contained in [1?
˙
P,1+
˙
P], so that
|S(h;
˙
P
0)|≤3·2
˙
P, and|S(h;
˙
P)−3|≤9
˙
P.
7. Use the fact that
β
n+1
i=1
kifi=(
β
n
i=1
kifi)+k n+1fn+1.
8. Since−M≤f(x)≤Mforx∈[a, b], Theorem 7.1.5(c) implies that we have
−M(b−a)≤

b
a
f≤M(b−a) whence the inequality follows.
9. Givenε>0 there existsδ
ε>0 such that if
˙
P<δ εthen|S(f;
˙
P)−

b
a
f|<ε.
Since
˙
P
n→0, there existsK εsuch that ifn>K εthen
˙
P n<δε, whence
|S(f;
˙
P
n)−

b
a
f|<ε. Therefore,

b
a
f= limnS(f;
˙
P n).
10. Sincegis not bounded, it is not Riemann integrable. Let
˙
P
nbe the partition
of [0, 1] intonequal subintervals with tags at the left endpoints, which are
rational numbers.
11. Iff∈R[a, b], then Exercise 9 implies that both sequences of Riemann sums
converge to

b
a
f.
12. LetP
nbe the partition of [0, 1] intonequal parts. If
˙
P nis this partition with
rational tags, thenS(f;
˙
P
n) = 1, while if
˙
Q nis this partition with irrational
tags, thenS(f;
˙
Q
n)=0.
13. If
˙
P<δ
ε:=ε/4α, then the union of the subintervals in
˙
Pwith tags in [c, d]
contains the interval [c+δ
ε,d−δ ε] and is contained in [c−δ ε,d+δ ε]. There-
foreα(d−c−2δ
ε)≤S(ϕ;
˙
P)≤α(d−c+2δ ε), whence|S(ϕ;
˙
P)−α(d−c)|≤
2αδ
ε<ε.
14. (a) Since 0≤x
i−1<xi, we have 0≤x
2
i−1
<xixi−1<x
2
i
, so that 3x
2
i−1
<x
2
i
+
x
ixi−1+x
2
i
=3q
2
i
<3x
2
i
. Therefore 0≤x i−1≤qi≤xi.

Chapter 7 — The Riemann Integral 53
(b) In fact, (x
2
i
+xixi−1+x
2
i−1
)·(x i−xi−1)=x
3
i
−x
3
i−1
.
(c) The terms inS(Q;
˙
Q) telescope.
(d) If
˙
Phas the tagst
iand
˙
P<δ, then|t i−qi|<δso that we have
|S(Q;
˙
P)−S(Q;
˙
Q)|<δ(b−a).
15. Let
˙
P={([x
i−1,xi],ti)}
n
i=1
be a tagged partition of [a, b] and let
˙
Q:=
{([x
i−1+c, xi+c],ti+c)}
n
i=1
so that
˙
Qis a tagged partition of [a+c, b+c]
and
˙
Q=
˙
P.Moreover,S(g;
˙
Q)=S(f;
˙
P) so that|S(g;
˙
Q)−

b
a
f|=
|S(f;
˙
P)−

b
a
f|<εwhen
˙
Q<δ ε.
Section 7.2
The Cauchy Criterion follows the standard pattern. It is used to obtain the
Squeeze Theorem 7.2.3, which is the tool used in proving the important inte-
grability theorems 7.2.5, 7.2.7 and 7.2.8. The only “tricky” proof is that of the
Additivity Theorem 7.2.9, but that proof can be soft-pedaled since the validity of
the theorem will seem obvious to most students.
Sample Assignment: Exercises 1, 2, 7, 8, 11, 12, 15, 18.
Partial Solutions:
1. If the conditions in 7.2.2(b) is satisﬁed, we can takenη=1/nand obtain the
condition in the statement. Conversely, if the statement holds andη>0is
given, we can taken∈Nsuch that 1/n<ηto get the desired
˙
P
n,
˙
Qn.
2. If the tags are all rational, thenS(h;
˙
P)≥1, while if the tags are all irrational,
thenS(h;
˙
P)=0.
3. Let
˙
P
nbe the partition of [0, 1] intonequal subintervals witht 1=1/nand
˙
Q
nbe the same subintervals tagged by irrational points.
4. No. Letf(x):=xifxis rational andf(x):=0 ifxis irrational in [0, 1].
There is no squeeze; that is,

b
a
(ωε−αε) is not small.
5. Ifc
1,...,cnare the distinct values taken byϕ, thenϕ
−1
(cj) is the union of a
ﬁnite collection{J
j1,...,Jjrj}of disjoint subintervals of [a, b]. We can write
ϕ=
β
n
j=1
β
rj
k=1
cjϕJjk.
6. Not necessarily. The Dirichlet function takes on only two values, butQ∩
[0,1] and [0,1]\Qare not intervals.
7. If
˙
P={([x
i−1,xi],ti)}
n
i=1
, takeϕ(x):=f(t i) forx∈[x i−1,xi) andϕ(b):=0,
so thatϕis a simple function. By the formula in Theorem 7.2.5, we have

b
a
ϕ=
β
n
i=1
f(ti)(xi−xi−1)=S(f;
˙
P).
8. Iff(c)>0 for somec∈(a, b), there existsδ>0 such thatf(x)>
1
2
f(c) for
|x−c|≤δ. Then

b
a
f≥

c+δ
c−δ
f≥
1
2
f(c)(2δ)>0. Ifcis an endpoint, a similar
argument applies.
9. The functionf(0) := 1 andf(x) := 0 elsewhere on [0,1] has integral 0. More
dramatically, consider Thomae’s function in Example 7.1.7.

54 Bartle and Sherbert
10. Leth:=f−gso thathis continuous. By Bolzano’s Theorem 5.3.7, ifhis
never 0, then eitherh(x)>0, orh(x)<0 for allx∈[a, b]. In the ﬁrst case
there existsγ>0 such thath(x)≥γ, whence

b
a
h≥γ(b−a)>0.
11. Sinceα
c(x)=f(x) forx∈[c, b], thenα c∈R[c, b]; similarlyω c∈R[c, b]. The
Additivity Theorem 7.2.9 implies thatα
candω care inR[a, b]. Moreover,

b
a
(ωc−αc)=2M(c−a)<εwhenc−a<ε/2M. The Squeeze Theorem 7.2.3
implies thatf∈R[a, b]. Further,|

b
a
f−

b
c
f|=|

c
a
f|≤M(c−a).
12. Indeed,|g(x)|≤1 for allx∈[0,1]. Sincegis continuous on every interval
[c,1] where 0<c<1, it belongs toR[c,1] and the preceding exercise applies.
13. Letf(x):=1/xforx∈(0,1] andf(0) := 0. Thenf∈R[c,1] for everyc∈(0,1),
butf/∈R[0,1] sincefis not bounded.
14. Use Mathematical Induction.
15. SupposeE={a=c
0<c1<···<c m=b}. Sincefif continuous on the inter-
val (c
i−1,ci), a two-sided version of Exercise 11 implies that its restriction is
inR[c
i−1,ci]. The preceding exercise implies thatf∈R[a, b]. The case where
aorbis not inEis similar.
16. Letm:= inff(x) andM:= supf(x). By Theorem 7.1.5(c), we have
m(b−a)≤

b
a
f≤M(b−a). By Bolzano’s Theorem 5.3.7, there existsc∈[a, b]
such thatf(c)=(

b
a
f)/(b−a).
17. Sinceg(x)>0, we havemg(x)≤f(x)g(x)≤Mg(x) for allx∈[a, b], whence
m

b
a
g≤

b
a
fg≤M

b
a
g. Since

b
a
g>0 (why?), Bolzano’s Theorem 5.3.7
implies that there existsc∈[a, b] such thatf(c)=(

b
a
fg)/(

b
a
g).
18. LetM:= supfand letp∈[a, b] be such thatf(p)=M. Givenε>0 there
exists an interval [c, d] containingpwithd−c>0 such thatM−ε≤f(x)≤M
forx∈[c, d]. Therefore
(M−ε)
n
(d−c)≤

d
c
f
n
≤

b
a
f
n
≤M
n
(b−a).
If we take thenth root, we have (M−ε)(d−c)
1/n
≤Mn≤M(b−a)
1/n
.Now
use the fact thatα
1/n
→1 forα>0 to complete the details.
19. The Additivity Theorem implies that the restrictions offto [−a,0] and [0,a]
are Riemann integrable. Let
˙
P
nbe a sequence of tagged partitions of [0,a]
with
˙
P
n→0 and let
˙
P
∗
n
be the corresponding “symmetric” partition of
[−a, a].
(a) Show thatS(f;
˙
P
∗
n
)=2S(f;
˙
P n)→2

a
0
f.
(b) Show thatS(f;
˙
P
∗
n
)=0.
20. Note thatx →f(x
2
) is an even continuous function.

Chapter 7 — The Riemann Integral 55
Section 7.3
The main results are the Fundamental Theorems, given in 7.3.1 and 7.3.5, and
the Lebesgue Integrability Criterion, stated in 7.3.12. The First Form 7.3.1 allows
for a ﬁnite setEwhere the functionFmay not be diﬀerentiable. It is useful to
point out that ifE=∅, then one does not need to assume thatFis continuous
at every point of [a, b]. However, one often encounters functions whereFis not
diﬀerentiable at every point (for exampleF(x)=
√
x). It is also worth stressing
that the hypothesis 7.3.1(c) isessential. The Second Form 7.3.5 is complementary
to the First Form, but is nowhere nearly as important in most situations.
The notion of a null set is an important one. No doubt most students will
think of countable sets, but it is worth pointing out that there are uncountable
null sets; however, it may be best to wait until the students encounter the Cantor
set in Section 11.2 before too much is made of this fact. Similarly, the proof of
the Lebesgue Criterion is given in Appendix C, but it is probably too complicated
for the average student at this level.
Sample Assignment: Exercises 2, 3, 5, 7, 9, 13, 18(a,c).
Partial Solutions:
1. Suppose thatE:={a=c
0<c1<···<c m=b}contains the points in [a, b]
where the derivativeF
ﬃ
(x) either does not exist, or does not equalf(x). If
we apply the proof of the 7.3.1 to [c
i−1,ci], we have thatf∈R[c i−1,ci] and

ci
ci−1
f=F(c i)−F(c i−1). Exercise 7.2.14 and Corollary 7.2.10 imply that
f∈R[a, b] and that

b
a
f=
β
m
i=1
(F(c i)−F(c i−1)) =F(b)−F(a).
2. We note thatH
nis continuous on [a, b] andH
ﬃ
n
(x)=x
n
for allx∈[a, b], so

b
a
x
n
dx=H n(b)−H n(a). HereE=∅.
3. LetE:={−1,1}.Ifx/∈E, the Chain Rule 6.1.6 implies thatG
ﬃ
(x)=
1
2
sgn(x
2
−1)·2x=xsgn(x
2
−1) =g(x). Alsog∈R[−2,3].
4. Indeed,B
ﬃ
(x)=|x|for allx.
5. (a) We have Φ
ﬃ
C
(x)=Φ
ﬃ
(x)=f(x) for allx∈[a, b], so Φ Cis also an antideriva-
tive offon [a, b].
6. By Theorem 7.2.13, we haveF
a(z)=

c
a
f+

z
c
f, so thatF c=Fa−

c
a
f.
7. Lethbe Thomae’s function. There is no functionH:[0,1]→Rsuch that
H
ﬃ
(x)=h(x) forxin some nondegenerate open interval; otherwise Darboux’s
Theorem 6.2.12 would be contradicted on this interval. So a ﬁnite setEwill
not suﬃce for this function.
8. Note thatF(0)=0= lim
x→0+
F(x) and that ifn∈N, then
lim
x→n−
F(x)=(n−1)n/2=F(n) = lim
x→n+
F(x).

56 Bartle and Sherbert
Therefore,Fis continuous forx≥0. AlsoF
ﬃ
(x)=n−1=[[x]] forxin
(n−1,n),n∈N. However,Fdoes not have (a two-sided) derivative at
n=0,1,2,.... Since there are only a ﬁnite number of these points in [a, b],
the Fundamental Theorem 7.3.1 implies that

b
a
[[x]]dx=F(b)−F(a).
9. (a)G(x)=F(x)−F(c), (b)H(x)=F(b)−F(x), (c)S(x)=F(sinx)−F(x).
10. IfF(x):=

x
a
f, then sincefis continuous on [a, b], Theorem 7.3.6 implies
thatF
ﬃ
(x)=f(x) for allx∈[a, b]. SinceG(x)=F(v(x)), the statement fol-
lows from the Chain Rule 6.1.6.
11. (a)F
ﬃ
(x)=2x(1 +x
6
)
−1
, (b)F
ﬃ
(x)=(1+x
2
)
1/2
−2x(1 +x
4
)
1/2
.
12.F(x):=x
2
/2 for 0≤x<1,F(x):=x−1/2 for 1≤x<2, andF(x):=
(x
2
−1)/2 for 2≤x≤3. Ifx = 2 thenF
ﬃ
(x)=f(x), butF
ﬃ
(2) does not exist.
13. For 0≤x≤2, we haveG(x)=

x
0
(−1)dt=−x; and for 2≤x≤3, we have
G(x)=

2
0
(−1)dt+

x
2
1dt=−2+(x−2) =x−4.G(x) is not diﬀerentiable
atx=2.
14. The Fundamental Theorem implies that iff
ﬃ
(x)≤2 for 0≤x≤2, thenf(x)−
f(0) =

x
0
f
ﬃ
(x)dx≤

x
0
2dx=2x, so thatf(x)≤2x+f(0)=2x−1 for
0≤x≤2. Thenf(2)≤3 so thatf(2) = 4 is impossible.
15. Sinceg(x)=

x+c
0
f−

x−c
0
fandfis continuous, theng
ﬃ
(x)=f(x+c)−
f(x−c).
16. IfF(x):=

x
0
f=−

x
1
f, thenF
ﬃ
(x)=f(x)=−f(x), so that 2f(x)=0 and
hencef(x) = 0 for allx∈[0,1].
18. (a) Takeϕ(t)=1+t
2
to get
1
2

t=1
t=0
(ϕ(t))
1/2
·ϕ
ﬃ
(t)dt=
12

x=2
x=1
x
1/2
dx=
13
x
3/2

2
1
=
1
3
(2
3/2
−1).
(b) Takeϕ(t)=1+t
3
to get
1
3

t=2
t=0
(ϕ(t))
−1/2
·ϕ
ﬃ
(t)dt=
13

x=9
x=1
x
−1/2
dx=
23
x
1/2

9
1
=
2
3
(9
1/2
−1) =
4
3
.
(c) Takeϕ(t)=1+
√
tto get 2

t=4
t=1
(ϕ(t))
1/2
·ϕ
ﬃ
(t)dt=2

x=3
x=2
x
1/2
dx=
4
3
x
3/2

3
2
=
4
3
(3
3/2
−2
3/2
).
(d) Takeϕ(t)=t
1/2
to get 2

t=4
t=1
cos(ϕ(t))·ϕ
ﬃ
(t)dt=2

x=2
x=1
cosxdx=
2(sin 2−sin 1).
19. In (a)–(c)ϕ
ﬃ
(0) does not exist. For (a), one can integrate over [c,4] and let
c→0+. For (b) the integrand is not bounded near 0, so the integral does
not exist. For (c), note that the integrand is even, so the integral equals
2

1
0
(1 +t)
1/2
dt. For (d),ϕ
ﬃ
(1) does not exist, so integrate over [0,c] and let
c→1−.
20. (b) It is clear that
ﬀ
n
Znis contained in
ﬀ
n,k
J
n
k
and that the sum of the
lengths of these intervals is≤
β
n
ε/2
n
=ε.
21. (a) The Product Theorem 7.3.16 implies that (tf±g)
2
≥0 is integrable.
(b) We have∓2t

b
a
fg≤t
2

b
a
f
2
+

b
a
g
2
. Now divide bytto obtain (b).

Chapter 7 — The Riemann Integral 57
(c) Lett→∞in (b).
(d) If

b
a
f
2
= 0, lett=(

b
a
g
2
/

b
a
f
2
)
1/2
in (b). Now replacefandgby|f|
and|g|.
22. Note that the composite function sgn◦his Dirichlet’s function, which is not
Riemann integrable.
Section 7.4
This optional section presents an alternative approach to the integral developed
by Gaston Darboux. Instead of Riemann sums using tags, this approach employs
upper and lower sums using suprema and inﬁma. The material in this section is
independent of the earlier sections of the chapter until the equivalence of the two
approaches is discussed. Because of time pressure, instructors will need to make
decisions about selection of material and this section provides an option.
Sample Assignment: Exercises 1, 4, 7, 9, 10, 12, 13, 14.
Partial Solutions:
1. (a)L(f;P
1)=(0+0+1)·1=1,U(f;P 1)=(1+1+2)·1=4.
(b)L(f;P
2)=(1/2+0+0+1/2+1+3/2)·
1
2
=7/4,
U(f;P
2)=(1+1/2+1/2+1+3/2+2)·(1/2)=13/4.
2. IfP=(a, x
2.,x3,...,xn−1,b) is any partition of [a, b] andf(x)=cfor all
x, thenL(f;P)=U(f;P)=c((x
2−a)+(x 3−x2)+(x 4−x3)+···+
(b−x
n−1)) =c(b−a).
3. IfPis a partition, then inf{f(x):x∈I
k}≤inf{g(x):x∈I k}for eachk,
so thatL(f;P)≤L(g;P). SincePis an arbitrary partition, we haveL(f)≤
L(g).
4. Ifk>0, then inf{kf(x):x∈I
j}=kinf{f(x):x∈I j}, whenceL(kf;P)=
kL(f;P). It follows thatL(kf)=kL(f).
5. It follows from Exercise 3 thatL(f)≤L(g)≤L(h) andU(f)≤U(g)≤U(h).
But ifL(f)=U(f)=AandL(h)=U(h)=A, it follows thatL(g)=A=U(g),
whencegis Darboux integrable with integralA.
6. Givenε>0, consider the partitionP
ε=(0,1−ε/2,1+ε/2,2). Then
U(f;P
ε)=2 andL(f;P ε)=2−ε. It follows that the integral is equal to 2.
7. (a) IfP
ε=(0,1/2−ε,1/2+ε,1), thenL(g;P ε)=1/2−εandU(g;P ε)=
1/2+ε.
(b) HereL(g;P
ε)=1/2−εandU(g;P ε)=1/2+13ε.

58 Bartle and Sherbert
8. If for somec∈Iwe havef(c)>0, then (by Theorem 4.2.9) there existsδ>0
such thatf(x)>f(c)/2>0 for|x−c|<δ,x∈I. Thus for some partitionP
c,
we haveL(f;P
c)>0, and thereforeL(f)>0.
9. Givenε>0, letP
1andP 2be partitions ofIsuch thatL(f j)−ε/2<L(f j;Pj),
and letP
ε=P1∪P2so thatL(f j)−ε/2<L(f j;Pε) forj=1,2. IfI 1,...,Im
are the subintervals ofIarising fromP ε, then it follows that
inf{f
1(x):x∈I k}+ inf{f 2(x):x∈I k)≤inf{f 1(x)+f 2(x):x∈I k}.
Thus we obtainL(f
1;Pε)+L(f 2;Pε)≤L(f 1+f2;Pε)≤L(f 1+f2).Hence
L(f
1)+L(f 2)−ε≤L(f 1+f2), whereε>0 is arbitrary.
10. Letf
1be the Dirichlet function (see Example 7.4.7(d)) and letf 2=1−f 1.
ThenL(f
1)=L(f 2)=0, butL(f 1+f2)=1.
11. If|f(x)|≤Mforx∈[a, b] andε>0, letP
εbe a partition such that the total
length of the subintervals that contain any of the pointsc
1,c2,...,cnis less
thanε/M. ThenU(f;P
n)−L(f;P n)<ε, so the Integrability Criterion 7.4.8
applies. Also 0≤U(f;P
n)≤ε, so thatU(f)=0.
12.L(f;P
n)=(0
2
+1
2
+···+(n−1)
2
)/n
3
=(n−1)n(2n−1)/6n
3
=
1
3
⊇
1−
3
2n
+
1
2n
2
∞
U(f;P
n)=(1
2
+2
2
+···+n
2
)/n
3
=n(n+ 1)(2n+1)/6n
3
=
1
3
π
1+
3
2n
+
1
2n
2
α
Therefore, 1/3=sup{L(f;P
n):n∈N}≤L(f)≤U(f)≤inf{U(f;P n):n∈N}=
1/3, and we conclude thatL(f)=U(f)=1/3.
13. It follows from Lemma 7.4.2 that ifPis a reﬁnement ofP
ε, thenL(f;P ε)≤
L(f;P) andU(f;P)≤U(f;P
ε), so thatU(f;P)−L(f;P)≤U(f;P ε)−
L(f;P
ε).
14. (a) By the Uniform Continuity Theorem 5.4.3,fis uniformly continuous on
I. Therefore ifε>0 is given, there existsδ>0 such that ifu, vinIand
|u−v|<δ, then|f(u)−f(v)|<ε/(b−a). Letnbe such thatn>(b−a)/δ
andP
n=(x0,x1,...,xn) be the partition ofIintonequal parts so that
x
k−xk−1=(b−a)/n<δ. Applying the Maximum-Minimum Theorem 5.3.4
to each subinterval, we getu
k,vkinIkso thatf(u k)=M kandf(v k)=m k.
ThenM
k−mk=f(u k)−f(v k)<ε/(b−a). Then 0≤U(f;P n)−L(f;P n)=
β
n
k=1
(Mk−mk)(xk−xk−1)≤ε. Sinceε>0 is arbitrary, it follows from
Corollary 7.4.9 thatfis integrable onI.

Chapter 7 — The Riemann Integral 59
(b) Iffis increasing onI, letP
nbe the partition as in (a). Thenf(x k)=M k
andf(x k−1)=m k. Then we have the “telescoping” sum
n
⊂
k=1
(Mk−mk)(xk−xk−1)=
b−a
n
n
⊂
k=1
(f(xk)−f(x k−1))
=
b−a
n
(f(x
1)−f(x 0)+f(x 2)−f(x 1)
+···+f(x
n)−f(x n−1))
=
b−an
(f(b)−f(a)).
Now givenε>0, choosen>(b−a)(f(b)−f(a))/ε. Then for partitionP
n,
we getU(f;P
n)−L(f;P n)=
n
⊂
k=1
(Mk−mk)(xk−xk−1)≤ε. Corollary 7.4.9
implies thatfis integrable onI.
15. We have 0≤U(f;P
n)−L(f;P n)≤K(b−a)
2
/n, and therefore
0≤U(f;P
n)−
b
a
f≤K(b−a)
2
/n.
Section 7.5
The proofs of the error estimates for the Trapezoidal, Midpoint and Simpson
formulas involve application of the Mean Value Theorem and the Bolzano Inter-
mediate Value Theorem. Since they are not particularly instructive, they are given
in Appendix D and the instructor may not wish to discuss them. Consequently,
it should be possible to cover this section in a single lesson.
Attention should be paid to the fact that, in the presence of convexity (or
concavity) of the integrand, one has bounds for the error in the Trapezoidal and
Midpoint Rules without examining the second derivative of the integrand.
Sample Assignment: Exercises 1, 2, 7, 8, 9, 17.
Partial Solutions:
1. Use (4) withn=4,a=1,b=2,h=1/4.Here 1/4≤f
ﬃﬃ
(c)≤2, soT 4≈0.697 02.
2. Use (10) withn=4,a=1,b=2,h=1/4.Sincef
(4)
(x)=24/x
5
, we have
3/4≤f
(4)
(c)≤2.HereS 4≈0.693 25.
3.T
4≈0.782 79.
4. The indexnmust satisfy 2/12n
2
<10
−6
; hencen>1000/
√
6≈408.25.
5.S
4≈0785 39.
6. The indexnmust satisfy 96/180n
4
<10
−6
; hencen≥28.

60 Bartle and Sherbert
7. Note thatp
(4)
(x) = 0 for allx.
8. Use the fact thatf
ﬃﬃ
(x)≥0 in (4) and (7). Geometrically the inequality is
reasonable because, if the function is convex, then the chord of the trapezoid
lies above the tangent to the graph. Iff
ﬃﬃ
(x)≤0, then the graph is concave
and the inequality is reversed.
9. A direct calculation.
10. A direct calculation.
11. Use Exercise 10.
12. The integral is equal to the area of one quarter of the unit circle. The error
estimates cannot be used because the derivatives ofhare unbounded on [0, 1].
Sinceh
ﬃﬃ
(x)≤0, the inequality isT n(h)<π/4<M n(h). See Exercise 8.
13. InterpretKas an area. Show thath
ﬃﬃ
(x)=−(1−x
2
)
3/2
and thath
(4)
(x)=
−3(1+4x
2
)(1 +x
2
)
−7/2
. To eight decimal places,π=3.141 592 65.
14. Approximately 3.653 484 49.
15. Approximately 4.821 159 32.
16. Approximately 0.835 648 85.
17. Approximately 1.851 937 05.
18. 1.
19. Approximately 1.198 140 23.
20. Approximately 0.904 524 24.

CHAPTER 8
SEQUENCES OF FUNCTIONS
In this chapter we study the pointwise and uniform convergence of sequences
of functions, so it draws freely from results in Chapter 3. After introducing
these concepts in Section 8.1, we show in Section 8.2 that one can interchange
certain important limiting operations (e.g., diﬀerentiation and integration) when
the convergence is uniform. Both of these sections are important and should be
discussed in detail.
Section 8.3 and 8.4 and more special. In Section 8.3 we use the results in
Section 8.2 to establish the exponential function on a ﬁrm foundation, after which
the logarithm is treated. In Section 8.4 we do the same for the sine and cosine
functions. Most of these properties will be familiar to the students, although the
approach will surely be new to them. A detailed discussion of Section 8.3 and 8.4
can be omitted if time is short.
Section 8.1
The distinction between ordinary (= pointwise) convergence and uniform conver-
gence of a sequence of functions on a setAis a subtle one. It centers on whether
the indexK(ε, x) can be chosen to be independent of the pointx∈A; that is,
whether the set{K(ε, x):x∈A}is bounded inR. If so, we can takeK(ε)tobe
the supremum of this set. However, it is not always easy to determine whether
this set is bounded for eachε>0. Often it is easier to obtain estimates for the
uniform norms introduced in Deﬁnition 8.1.7.
Sample Assignment: Exercises 1, 2, 3, 4, 7, 11, 12, 13, 14, 17.
Partial Solutions:
1. Note that 0≤f
n(x)≤x/n→0asn→∞.
2. Note thatf
n(0) = 0 for alln.Ifx>0, we have|f n(x)|≤1/(nx)→0as
n→∞.
3. Note thatf
n(0) = 0 for alln∈N.Ifx>0, then|f n(x)−1|<1/(nx)→0as
n→∞.
4. Ifx∈[0,1), then|f
n(x)|≤x
n
→0. Ifx= 1, thenf n(1)=1/2 for alln∈N.
Ifx>1, then|f
n(x)−1/=1/(1 +x
n
)≤(1/x)
n
→0.
5. Note thatf
n(0) = 0 for alln.Ifx>0, then|f n(x)|≤1/(nx)→0.
6. Note thatf
n(0) = 0 for alln.If0<ε<π/2, letM ε:= tan(π/2−ε)>0
so that ify>M
ε, thenπ/2−ε<Arctany<π/2. Therefore ifn>M ε/x,
thenπ/2−ε<Arctannx<π/2. Similarly ifx<0.
61

62 Bartle and Sherbert
7. Note thatf
n(0) = 1 for alln.Ifx>0, then 0<e
−x
<1 so that 0≤e
−nx
=
(e
−x
)
n
→0.
8. Note thatf
n(0) = 0 for alln.Ifx>0, it follows from Exercise 7 that 0≤
x(e
−x
)
n
→x·0=0.
9. For both functionsf
n(0) = 0 for alln.Ifx>0, then 0≤x
2
e
−nx
=x
2
(e
−x
)
n
→
0, since 0<e
−x
<1. For the second function, use Theorem 3.2.11 and
[(n+1)
2
x
2
e
−(n+1)x
]/[n
2
x
2
e
−nx
]=(1+1/n)
2
e
−x
→e
−x
<1.
10. Ifx∈Z, then cosπx=±1, so that (cosπx)
2
= 1 and the limit equals 1. If
x/∈Z, then 0≤(cosπx)
2
<1 and the limit equals 0.
11. Ifx∈[0,a], then|f
n(x)|≤a/n. However,f n(n)=1/2.
12. Ifx∈[a,∞), then|f
n(x)|≤1/(na). However,f n(1/n)=1/2.
13. Ifa>0, then|f
n(x)−1|≤1/(na)on[a,∞). However,f n(1/n)=1/2.
14. Ifx∈[0,b], then|f
n(x)|≤b
n
. However,f n(2
−1/n
)=1/3.
15. Ifx∈[a,∞), then|f
n(x)|≤1/(na). However,f n(1/n)=
1
2
sin 1>0.
16. If 0<ε<π/2, letM
ε:= tan(π/2−ε)>0, so that ifna≥M ε, thenπ/2−ε≤
Arctanna<π/2. Hence, ifx≥aandn≥M
ε/a, thennx≥M εandπ/2−ε≤
Arctannx<π/2. However,f
n(1/n) = Arctan 1 =π/4>0.
17. Ifx∈[a,∞), then|f
n(x)|≤(e
−a
)
n
. However,f n(1/n)=1/e.
18. The maximum off
non [0,∞)isatx=1/n,sof n
[0,∞)=1/(ne).
19. The maximum off
non [0,∞)isatx=2/n,sof n
[0,∞)=4/(ne)
2
.
20. Ifnis suﬃciently large, the maximum off
non [a,∞)isatx=a>0, so that
f
n
[a,∞)=n
2
a
2
/e
na
→0. However,f n
[0,∞)=fn(2/n)=4/e
2
.
21. Givenε>0, letK
1(ε/2) be such that ifn≥K 1(ε/2) andx∈A, then
|f
n(x)−f(x)|<ε/2; also letK 2(ε/2) be such that ifn≥K 2(ε/2) and
x∈A, then|g
n(x)−g(x)|<ε/2. LetK 3:= sup{K 1(ε/2),K 2(ε/2)}so that
ifn≥K
3andx∈A, then|(f n+gn)(x)−(f+g)(x)|≤|f n(x)−f(x)|+
|g
n(x)−g(x)|<ε.
22. We have|f
n(x)−f(x)|=1/nfor allx∈R. Hence (f n) converges uniformly on
Rtof. However,|f
2
n
(n)−f
2
(n)|≥2 so that (f
2
n
) does not converge uniformly
onRtof.
23. LetMbe a bound for (f
n(x)) and (g n(x)) onA, whence also|f(x)|≤M.
The Triangle Inequality gives|f
n(x)gn(x)−f(x)g(x)|≤M[|f n(x)−f(x)|+
|g
n(x)−g(x)|] forx∈A.
24. Sincegis uniformly continuous on [−M,M], givenε>0 there existsδ
ε>0
such that if|u−v|<δ
εandu, v∈A, then|g(u)−g(v)|<ε.If(f n) converges
uniformly tofonA, givenδ>0 there existsK(δ) such that ifn≥K(δ)
andx∈A, then|f
n(x)−f(x)|<δ. Therefore, ifn≥K(δ ε) andx∈A, then
|g(f
n(x))−g(f(x))|<ε.

Chapter 8 — Sequences of Functions 63
Section 8.2
The proof of Theorem 8.2.2 is short and understandable; it should be discussed
in detail. That of Theorem 8.2.3 is more delicate; observe that it depends on the
Mean Value Theorem 6.2.4 in two places. Note especially that the hypothesis in
Theorem 8.2.3 is that thesequence of derivativesis uniformly convergent (and
that the uniform convergence of the sequence of functions is a conclusion, rather
than a hypothesis). The only delicate part of the proof of Theorem 8.2.4 is to
show that the limit function is integrable. The Bounded Convergence Theorem
8.2.5 will be considerably strengthened in Section 10.4. Dini’s Theorem 8.2.6 is
interesting in that it shows that monotone convergence for continuous functions
to a continuous limit implies the uniformity of the convergence on [a, b].
Sample Assignment: Exercises 1, 2, 4, 5, 7, 8, 14, 16, 19.
Partial Solutions:
1. The limit function isf(x):=0 for 0≤x<1,f(1) := 1/2 andf(x):=1 for
1<x≤2. Since it is discontinuous, while thef
nare all continuous, the con-
vergence cannot be uniform.
2. The convergence is not uniform, becausef
n(1/n)=n, whilef(x) = 0 for all
x∈[0,1].
3. Letf
n(x):=1/nifxis rational andf n(x):=0 ifxis irrational.
4. Ifε>0 is given, letKbe such that ifn≥K, thenf
n−f I<ε/2. Then
|f
n(xn)−f(x 0)|≤|f n(xn)−f(x n)|+|f(x n)−f(x 0)|≤ε/2+|f(x n)−f(x 0)|.
Sincefis continuous (by Theorem 8.2.2) andx
n→x0, then|f(x n)−f(x 0)|<
ε/2 forn≥K
ﬃ
, so that|f n(xn)−f(x 0)|<εforn≥max{K, K
ﬃ
}.
5. Givenε>0, there existsδ>0 such that ifx, u∈Rand|x−u|<δ, then
|f(x)−f(u)|<ε. Now require that 1/n<δ.
6. Heref(0) = 1 andf(x)=0 forx∈(0,1]. Since thef
nare continuous on
[0, 1] butfis not, the convergence cannot be uniform on [0, 1]. Alternatively,
note thatf
n(1/n)→1/e.
7. Givenε:= 1, there existsK>0 such that ifn≥Kandx∈A, then|f
n(x)−
f(x)|<1, so that|f
n(x)|≤|f K(x)|+ 1 for allx∈A.IfM:= max{f 1A,...,
f
K−1A,fKA+1}, then|f n(x)|≤Mfor alln∈N,x∈A. Therefore
|f(x)|≤Mfor allx∈A.
8. Since 0≤f
n(x)≤nx≤non [0, 1] and 0≤f n(x)≤1/x≤1on[1,∞), we have
0≤f
n(x)≤non [0,∞). Moreover,f(x) := lim(f n(x)) = 0 forx= 0 and
f(x)=1/xforx>0, which is not bounded on [0,∞). Sincefis not con-
tinuous on [0,∞), the convergence is not uniform. Alternatively,f
n(1/
√
n)=
√
n/2.

64 Bartle and Sherbert
9. Sincef(x) = 0 for allx∈[0,1], we havef
ﬃ
(1) = 0. Alsof n−f
[0,1]≤1/n,
so the convergence of (f
n) is uniform. Alsog(x) = lim(x
n−1
) so thatg(1)=1.
The convergence of the sequence of derivatives is not uniform on [0,1].
10. Hereg
n
[0,∞)≤1/nso (g n) converges uniformly to the zero function. How-
ever, lim(g
ﬃ
n
(x)) =−1 forx= 0, and = 0 forx>0. Hence (limg n)
ﬃ
(0)=0 =
1 = lim(g
ﬃ
n
(0)).The sequence (g
ﬃ
n
) does not converge uniformly.
11. The Fundamental Theorem 7.3.1 implies that

x
a
f
ﬃ
n
=fn(x)−f n(a). Now
apply Theorem 8.2.4.
12. The functionf
n(x):=e
−nx
2
is decreasing on [1, 2] andf n
[1,2]=e
−n
. Hence
Theorem 8.2.4 can be applied.
13. Ifa>0, thenf
n
[a,π]≤1/(na) and Theorem 8.2.4 applies. On the interval
[0,π] the limit function isf(0) := 1 andf(x):=0 forx∈(0,π]. Moreover
f
n
[0,π]= 1. Hence it follows from Theorem 8.2.5 that

π
0
f= 0. This can
also be proved directly by changing the variablev=nxand estimating the
integrals.
14. The limit function isf(0) := 0,f(x):=1 forx∈(0,1] and the convergence is
not uniform on [0,1]. Her lim

1
0
fn= lim(1−(1/n) ln(n+ 1)) = 1 and

1
0
f=1.
15. Hereg
n(0) = 0 for alln, andg n(x)→0 forx∈(0,1] by Theorem 3.2.11. The
functiong
nis maximum on [0, 1] atx=1/(n+ 1), whenceg n
[0,1]≤1 for
alln. Now apply Theorem 8.2.5. Or evaluate the integrals directly.
16. Eachf
nis Riemann integrable since it has only a ﬁnite number of disconti-
nuities. (See Exercise 7.1.13 or the Lebesgue Integrability Criterion 7.3.12.)
17. Heref(x):=0 forx∈[0,1] and we havef
n−f
[0,1]=1.
18. Heref(x):=0 forx∈[0,1) andf(1) := 1 and we havef
n−f
[0,1]=1.
19. Heref(x) := 0 for allx∈[0,∞) and|f
n(n)−f(n)|=1.
20. Letf
n(x):=x
n
forx∈[0,1].
Section 8.3
There are many diﬀerent approaches to the exponential and logarithmic functions:
see R.G. Bartle’sElements of Real Analysis. The approach here is based on
obtaining the exponential function as a limit of polynomials (which, in fact, are the
partial sums of the Maclaurin series for the exponential function). The uniqueness
and basic properties of the exponential function are based on the diﬀerential
equation and initial conditions it satisﬁes.
The logarithm is obtained as the function that is inverse to the exponential
function. The power functionsx →a
α
and the functionsx →log
axare often
useful, but can largely be left to the student.

Chapter 8 — Sequences of Functions 65
Sample Assignment: Exercises 1, 2, 3, 5, 6, 8, 10, 13.
Partial Solutions:
1. To establish the inequality, letA:=x>0 and letm→∞in (5). For the
estimate one, takex= 1 andn= 3 to obtain|e−2
2
3
|<1/12, soe<2
3
4
.
Since (E
n(1)) is increasing, we also have 2
2
3
<e.
2. Note that ifn≥9, then 2/(n+ 1)!<6×10
−7
<5×10
−6
.Hencee≈2.71828.
3. EvidentlyE
n(x)≤e
x
for alln∈N,x≥0. To obtain the other inequality,
apply Taylor’s Theorem 6.4.1 to [0,a] and note that ifc∈[0,a], then 1≤
e
c
≤e
a
.
4. To obtain the inequality, replacenbyn+ 1 and takea= 1 in Exercise 3. Since
2<e<3, we havee/(n+ 1) forn≥2. Ife=m/n, thenen!−(1+1+···+
1/n!)n! is an integer in (0, 1), which is impossible.
5. Note that 0≤t
n
/(1 +t)≤t
n
fort∈[0,x].
6. ln 1.1≈0.0953 and ln 1.4≈0.3365.Taken>19,999.
7. ln 2≈0.6931. Note thate/2−1<0.36 and (0.36)
8
/8<0.000 04.
8. Iff(0) = 0, the argument in 8.3.4 show thatf(x) = 0 for allx,sowe
takeK=0. Iff(0) = 0, theng(x):=f(x)/f(0) is such thatg
ﬃ
(x)=g(x)
for allxandg(0) = 1. It follows from 8.3.4 thatg(x)=E(x),whencef(x)=
f(1)e
x
.
9. Note that if the means are equal, then we must have 1 +x
k=E(x k) for all
k. It follows thatx
k= 0, whencea k=Afor allk.
10.L
ﬃ
(1) = lim[L(1+1/n)−L(1)]/(1/n) = limL((1+1/n)
n
)=L(lim(1+
1/n)
n
)=L(e)=1.
11. (a) SinceL(1)=0,wehave1
α
=E(αL(1)) =E(0)=1.
(b) This follows from the fact thatE(z)>0 for allz∈R.
(c) (xy)
a
=E(αL(xy)) =E(αL(x)+αL(y)) =E(αL(x))·E(αL(y)) =x
α
·y
α
.
(d) Since (1/y)
α
=E(αL(1/y)) =E(−αL(y))=(E(αL(y))
−1
=(y
α
)
−1
, the
statement follows from (c).
12. (a)x
α+β
=E((α+β)L(x)) =E(αL(x)+βL(x)) =E(αL(x))·E(βL(x)) =
x
α
·x
β
.
(b) (x
α
)
β
=E(βL(x
α
)) =E(βαL(x)) =x
αβ
, and similarly for (x
β
)
α
.
(c)x
−α
=E(−αL(x))=(E(αL(x))
−1
=(x
α
)
−1
=1/x
α
.
(d) Ifx>1, thenL(x)>0, so that ifα<β, thenαL(x)<βL(x). SinceEis
strictly increasing, we deduce thatx
α
=E(αL(x))<E(βL(x)) =x
β
.
13. (a) Ifα>0, it follows from 8.3.13 thatx →x
α
is strictly increasing. Since
lim
x→0+
L(x)=−∞, use 8.3.7 to show that lim
x→0+
x
α
=0.
(b) Ifα<0, thenαLis strictly decreasing, whencex →x
α
is strictly decreas-
ing. Here lim
x→0+
αL(x)=∞, so lim
x→0+
x
α
= lim
y→∞
E(y)=∞, and lim
x→∞
αL(x)=
−∞, so lim
x→∞
x
α
= lim
y→−∞
E(y)=0.

66 Bartle and Sherbert
14. By 8.3.14, ifx>0 anda>0,a = 1, then (log
ax)(lna)= lnx, whence
a
log
a
x
=E((log
ax)(lna)) =E(lnx)=xforx>0. Similarly, since ln(a
y
)=
ln(E(ylna)) =ylna,wehavelog
a(a
y
) = (lna
y
)/(lna)=yfor ally∈R.
15. Use 8.3.14 and 8.3.9(vii).
16. Use 8.3.14 and 8.3.9(viii).
17. Indeed, we have log
ax= (lnx)/(lna) = [(lnx)/(lnb)]·[(lnb)/(lna)] ifa =1,
b = 1. Now takea=10,b=e.
Section 8.4
Although the characterization of the sine and cosine functions given here is not the
traditional approach to these functions, it has several advantages. Indeed, the
most important properties of these functions follow quickly from the fact that
they satisfy the diﬀerential equationf
ﬃﬃ
(x)=−f(x) for allx∈R, and that any
function satisfying this diﬀerential equation is a linear combination of sin and cos.
[Other approaches to the trigonometric functions are sketched in R. G. Bartle’s
Elements of Real Analysis.]
Sample Assignment: Exercises 1, 2, 4, 6, 7, 8.
Partial Solutions:
1. Ifn>2|x|, then|cosx−C
n(x)|≤(16/15)|x|
2n
/(2n)!. Hence cos(0.2)≈
0.980 067 and cos 1≈0.549 302. As for the sine function, ifn>2|x|, then
|sinx−S
n(x)|≤(16/15)|x|
2n
/(2n)!. Hence sin(0.2)≈0.198 669 and sin 1≈
0.841 471.
2. It follows from Corollary 8.4.3 that|sinx|≤1 and|cosx|≤1.
3. Ifx<0, then−x>0 so property (vii) never holds. However, ifx<0, we
have−(−x)≤S(−x)=−S(x)≤−x, whence−|x|=x≤S(x)≤−x=|x|.It
follows from 8.4.8(ix) that−x
3
/6≤−(sinx−x)≤0ifx≥0. Hence, ifx<0,
we havex
3
/6≤−(sinx−x)≤0, whence|sinx−x|≤|x|
3
/6.
4. We integrate 8.4.8(x) twice on [0,x]. Note that the polynomial on the left
has a zero in the interval [1.56, 1.57], so 1.56≤π/2.
5. Exercise 8.4.4 shows thatC
4(x)≤cosx≤C 3(x) for allx∈R. Integrating sev-
eral times, we getS
4(x)≤sinx≤S 5(x) for allx>0. Show thatS 4(3.05)>0
andS
5(3.15)<0. (This procedure can be sharpened.)
6. Clearlys
n(0) = 0 andc n(0) = 1; alsos
ﬃ
n
(x)=c n(x) andc
ﬃ
n+1
(x)=s n(x) for
x∈R,n∈N. Show that if|x|≤Aandm>n>2A, then|c
m(x)−c n(x)|<
(16/15)A
2n
/(2n)!, whence it follows that|c(x)−c n(x)|<(16/15)A
2n
/(2n)!
and the convergence of (c
n)tocis uniform on each interval [−A, A]. Similarly
for (s
n). Sincec
ﬃﬃ
n+1
=cnands
ﬃﬃ
n+1
=snforn∈N, property (j) holds. Property
(jj) is evident, and it follows froms
ﬃ
n
=cnandc
ﬃ
n+1
=snthats
ﬃ
=candc
ﬃ
=s.

Chapter 8 — Sequences of Functions 67
7. Note that the derivativeD[(c(x))
2
−(s(x))
2
] = 0 for allx∈R. To establish
uniqueness, argue as in 8.4.4.
8. Letg(x):=f(0)c(x)+f
ﬃ
(0)s(x) forx∈R, so thatg
ﬃﬃ
(x)=g(x),g(0) =f(0)
andg
ﬃ
(0) =f
ﬃ
(0). Therefore the functionh(x):=f(x)−g(x) has the property
thath
ﬃﬃ
(x)=h(x) for allx∈Randh(0)=0,h
ﬃ
(0) = 0. Thus it follows as in
the proof of 8.4.4 thatg(x)=f(x) for allx∈R, so thatf(x)=f(0)c(x)+
f
ﬃ
(0)s(x). Now note thatf 1(x):=e
x
andf 2(x):=e
−x
satisfyf
ﬃﬃ
(x)=f(x)
forx∈R. Hencef
1(x)=c(x)+s(x) andf 2(x)=c(x)−s(x), whence it follows
thatc(x)=
1
2
(e
x
+e
−x
) ands(x)=
1
2
(e
x
−e
−x
).
9. Ifϕ(x):=c(−x), show thatϕ
ﬃﬃ
(x)=ϕ(x) andϕ(0)=1,ϕ
ﬃ
(0) = 0, so that
ϕ(x)=c(x) for allx∈R. Thereforecis even.
10. It follows from Exercise 8 thatc(x)>e
x
/2>0 for allx∈R. Therefores
is strictly increasing onRand, sinces(0) = 0, it follows thatcis strictly
increasing on (0,∞). Thus 1 =c(0)<c(x) for allx∈(0,∞); sincecis even,
we deduce thatc(x)≥1 for allx∈R. Since lim
x→∞
e
x
=∞and lim
x→∞
e
−x
=0,
it follows from Exercise 8 that lim
x→∞
c(x) = lim
x→∞
s(x)=∞.

CHAPTER 9
INFINITE SERIES
Students have been exposed to much of the material in this chapter in their
introductory calculus course; however, their recollection of this material probably
will not go much beyond the mechanical application of some of the “Tests”. At this
point, they should be prepared to approach the subject on a more sophisticated
level.
Instructors will recall that an introduction to series was given in Section 3.7
and it would be well to review that section very brieﬂy. Since Section 9.1 is quite
short it is possible to do that in one lesson. Although much of Section 9.2 will be
familiar, the short Section 9.3 will probably be new. Section 9.4 is an interesting
one, and ties this discussion together with Chapter 8.
Section 9.1
The notion of absolute convergence is rather subtle, and should be stressed. The
discussion about rearrangements will help the student to realize the importance
of absolute convergence. If the Cauchy Condensation Test in Exercise 3.7.15 has
not been discussed before, it should be covered now.
Sample Assignment: Exercises 1, 2, 4, 7, 8, 11, 12.
Partial Solutions:
1. Lets
nbe thenth partial sum of
β
∞
1
an, lett nbe thenth partial sum
of
β
∞
1
|an|, and suppose thata n≥0 forn>P.Ifm>n>P, show that
t
m−tn=sm−sn. Now apply the Cauchy Criterion.
2. Replace each strictly negative term by 0 to obtain
β
p
n, and replace each
strictly positive term by 0 to obtain
β
q
n.If
β
p nis convergent, then
β
q n
is also convergent sinceq n=an−pnand
β
a nis convergent (see Exercise
3.7.4). In this case,|a
n|=pn−qn, so that
β
a nis absolutely convergent, a
contradiction.
3. Take positive terms until the partial sum exceeds 1, then take negative terms
until the partial sum is less than 1, then take positive terms until the partial
sum exceeds 2, etc.
4. It was used together with the Cauchy Criterion to assure that givenε>0
there existsN<qsuch that
β
q
N+1
|xk|<ε.
5. Yes. LetM>0 be such that every partial sumt
nof
β
|a n|satisﬁes 0≤t n≤M.
If
β
b
kis rearrangement of
β
a nand ifu k:=|b1|+···+|b k|, then there
existsn∈Nsuch that every term|b
i|inu kis contained int nand hence
0≤u
k≤tn≤M. Therefore
β
b kis absolutely convergent.
68

Chapter 9 — Infinite Series 69
6. Use Mathematical Induction to show that ifn≥2, thens
n=−ln 2−lnn+
ln(n+ 1). Yes, sincea
n<0 for alln≥2.
7. (a) If|b
k|≤Mfor allk∈N, then|a nbn+···+a mbm|≤M(|a n|+···+|a m|).
Now apply the Cauchy Criterion 3.7.4.
(b) Letb
k:= +1 ifa k≥0 andb k:=−1ifa k<0. Then
β
a kbk=
β
|a k|.
8. Leta
k:= (−1)
k
/
√
k,soa
2
k
=1/k.
9. Sinces
2n−sn=an+1+···+a 2n≥na2n=
1
2
(2na2n), then lim(2na 2n)=0.
Similarlys
2n+1−sn≥(n+1)a 2n+1≥
1
2
(2n+1)a 2n+1, so that we have
lim(2n+1)a
2n+1= 0. Consequently lim(na n)=0.
10. Consider
β
∞
2
1/(nlnn), which diverges by Exercise 3.7.17(a).
11. Indeed, if|n
2
an|≤Mfor alln, then|a n|≤M/n
2
so Example 3.7.6(c) and the
Comparison Test 3.7.7(a) apply.
12. If 0<a<1, thena
n
→0, so 1/(1 +a
n
)→1 and the series diverges by the
nth Term Test 3.7.3. Similarly, ifa= 1, then 1/(1 +a
n
)=
1
2
.Ifa>1, then
1/(1 +a
n
)<(1/a)
n
and the series converges by comparison with a geometric
series with ratio 1/a <1.
13. (a) Rationalize to obtain
β
x
nwherex n:= [
√
n(
√
n+1+
√
n)]
−1
and note
thatx
n≈yn:= 1/(2n). Now apply the Limit Comparison Test 3.7.8 to show
the series diverges.
(b) Rationalize and compare with
β
1/n
3/2
to show the series converges.
14. If
β
a
nis absolutely convergent, then the partial sums (t n)of
β
|a n|are
bounded, say byM. It is evident that the absolute value of the partial sums
of any subseries ofa
nare also bounded byM, so these subseries are also
(absolutely) convergent.
Conversely, if every subseries of
β
a
nis convergent, then the subseries
consisting of the strictly positive (and strictly negative) terms are absolutely
convergent, whence it follows that
β
a
nis absolutely convergent.
15. If (i) exists, lets
n:=
β
∞
k=1
ck. For ﬁxedn∈N, choosei 0,j0such that
{c
1,...,cn}⊆{a ij:i≤i 0,j≤j 0}. Thens n≤
β
i0
i=1
β
j0
j=1
aij≤
β
∞
i=1
β
∞
j=1
aij=B. Sincen∈Nis arbitrary, it follows thatCexists and
C≤B.
If (ii) holds, givenn∈N, choosem∈Nsuch that{a
i1,...,ain}⊆
{c
1,...,cm}. Then
β
n
j=1
aij≤
β
m
k=1
ck≤Cso that
β
∞
j=1
aij≤Cfor all
j∈N. Now chooseNsuch that{a
ij:i≤m, j≤n}⊆{c k:k≤N}. Then
β
m
i=1
β
n
j=1
aij≤
β
N
k=1
ck≤C. First letn→∞, then letm→∞to get
B≤C.
16. Note that
β
∞
j=1
aij=−1ifi= 1 and = 0 ifi>1, whence
β
∞
i=1
β
∞
j=1
aij=
−1. On the other hand,
β
∞
i=1
aij=1 ifj= 1 and = 0 ifj>1, whence
β
∞
j=1
β
∞
i=1
aij=1.

70 Bartle and Sherbert
Section 9.2
The results presented in this section are primarily designed to test for absolute
convergence. All of these tests are very useful, but they are not deﬁnitive in
the sense that there are some series that do not yield to them, but require more
delicate tests (such as Kummer’s and Gauss’s Tests that are presented in more
advanced treatises).
Sample Assignment: Exercises 1, 2(a,b), 3(a,c,e), 5, 7(a,c), 9, 12(a,d), 16.
Partial Solutions:
1. (a) Convergent; compare with
β
1/n
2
.
(b) Divergent; apply 9.2.1 withb
n:= 1/n.
(c) Divergent; note that 2
1/n
→1.
(d) Convergent; apply 9.2.3 or 9.2.5.
2. (a) Divergent; apply 9.2.1 withb
n:= 1/n.
(b) Convergent; apply 3.7.7 or 9.2.1 withy
n:=n
−3/2
.
(c) Convergent; use 9.2.4 and note that (n/(n+ 1))
n
→1/e <1.
(d) Divergent; thenth term does not tend to 0.
3. (a) (lnn)
p
<nfor largen, by L’Hospital’s Rule.
(b) Convergent; apply 9.2.3.
(c) Convergent; note that (lnn)
lnn
>n
2
for largen. Now apply 3.7.7 or 9.2.1.
(d) Divergent; note that (lnn)
ln lnn
= exp((ln lnn)
2
)<exp(lnn)=nfor
largen. Now apply 3.7.7 or 9.2.1.
(e) Divergent; apply 9.2.6 or Exercise 3.7.15.
(f) Convergent; apply 9.2.6 or Exercise 3.7.15.
4. (a) Convergent; apply 9.2.2 or 9.2.4.
(b) Divergent; apply 9.2.4.
(c) Divergent; note thate
lnn
=n.
(d) Convergent; note that (lnn) exp(−n
1/2
)<nexp(−n
1/2
)<1/n
2
for
largen, by L’Hospital’s Rule.
(e) Divergent; apply 9.2.4.
(f) Divergent; apply 9.2.4.
5. Compare with
β
1/n
2
.
6. Apply the Integral Test 9.2.6.
7. (a,b) Convergent; apply 9.2.5.
(c) Divergent; note thatx
n≥(2/4)(4/6)···(2n/(2n+2))=1/(n+ 1). Or,
apply 9.2.9.
(d) Convergent; apply 9.2.9.
8. Here lim(x
1/n
n
)=a<1.
9. Ifm>n≥K, then|s
m−sn|≤|x n+1|+···+|x m|<r
n+1
/(1−r). Now let
m→∞.

Chapter 9 — Infinite Series 71
10. Relation (5) implies that|x
n+k|≤r
k
|xn|whenn≥K. Therefore, ifm>
n≥K, we have|s
m−sn|≤(r+r
2
+···+r
m−n
)|xn|<|x n|(r/(1−r)). Now
take the limit asm→∞.
11. Letm>n≥K. Use (12) to get (a−1)(|x
n+1|+···+|x m|)≤n|x n+1|−
m|x
m+1|, whence|s m−sn|≤|x n+1|+···+|x m|≤|x n+1|n/(a−1). Now take
the limit asm→∞.
12. (a) A crude estimate of the remainder is given bys−s
4=1/6·7+
1/7·8+···<

∞
5
x
−2
dx=1/5.Similarlys−s 10<1/11 ands−s n<1/(n+1),
so that 999 terms suﬃce to gets−s
999<1/1000. [In this case the series tele-
scopes and we haves
n=1/2−1/(n+ 2).]
(d) Ifn≥4, thenx
n+1/xn≤5/8 so (by Exercise 10)|s−s 4|≤5/12. Ifn≥10,
thenx
n+1/xn≤11/20 so that|s−s 10|≤(10/2
10
)(11/9)<0.012. Ifn= 14,
then|s−s
14|<0.000 99. Alternatively, ifn≥4, thenx
1/n
n
≤1/
√
2so(by
Exercise 9)|s−s
4|≤1/4(
√
2−1)<0.61. Ifn≥10, thenx
1/n
n
≤(1/2)(10)
1/10
so that|s−s 10|<0.017. Ifn= 15, then|s−s 15|<0.000 69.
13. (b) Here 1/5
√6+1/6
√
7+···<

∞
4
x
−3/2
dx= 1. Therefore we have
β
∞
n+1
<

∞
n
x
−3/2
dx=2/
√
n,so|s−s 10|<0.633 and|s−s n|<0.001 whenn>4×10
6
.
(c) Ifn≥4, then|s−s
n|≤(0.694)x nso that|s−s 4|<0.065. Ifn≥10, then
|s−s
n|≤(0.628)x nso that|s−s 10|<0.000 023.
14. Note thats
3n>1+1/4+1/7+···+1/(3n+ 1), which is not bounded.
15. Since lnn=

n
1
t
−1
dt <1/1+1/2+···+1/(n−1), it follows that 1/n<c n.
Sincec
n−cn+1= ln(n+1)−lnn−1/(n+1)=1/θ n−1/(n+ 1) by the Mean
Value Theorem, whereθ
n∈(n, n+ 1), we havec n−cn+1>0. Therefore the
decreasing sequence (c
n) converges, say toC. An elementary calculation
shows thatb
n=c2n−cn+ ln 2, so thatb n→ln 2.
16. Note that, for an integer withndigits, there are 9 ways of picking the ﬁrst
digit and 10 ways of picking each of the othern−1 digits. Thus there are 8
“sixless” valuesn
kfrom 1 to 9, there are 8·9 such values from 10 to 99, there
are 8·9
2
values between 100 and 999, and so on. Hence
β
1/n kis dominated
by 8/1+8·9/10+8·9
2
/10
2
+···= 80.
There is one value ofm
kfrom 1 to 9, there is one value from 10 to 19,
one from 20 to 29, etc. Hence the (grouped) terms of
β
1/m
kdominate
1/10+1/20 +···=(1/10)
β
1/k, which is divergent.
There are 9 values ofp
kfrom 1 to 9, there are 9 such values from 10 to
19, and so on. Hence the (grouped) terms of
β
1/p
kdominate 9(1/10) +
9(1/20) +···=(9/10)
β
1/k, which is divergent.
17. The terms are positive and lim(n(1−x
n+1/xn)) =q−p; therefore, it follows
from 9.2.9 that the series is convergent ifq>p+ 1 and is divergent ifq<p+1.
Ifq=p+ 1, use 9.2.1 withy
n:= 1/nto establish divergence.
18. Here lim(n(1−x
n+1/xn))=(c−a−b) + 1, so the series is convergent ifc>
a+band is divergent ifc<a+b. [Ifc=a+bandab >0, one can show

72 Bartle and Sherbert
thatx
n+1/xn≥n/(n+ 1) so that (nx n) is an increasing sequence, whence
the series is divergent. The restriction thatab≥0 can be removed by using
a stronger test, such as Kummer’s or Gauss’s test.]
19. Hereb
1+b2+···+b n=A
1/2
−(A−A n)
1/2
→A
1/2
, so that
β
b nconverges
toA
1/2
. Alsob n>0 anda n/bn=(A−A n−1)
1/2
+(A−A n)
1/2
→0.
20. Here (b
n) is a decreasing sequence converging to 0 andb 1+b2+···+b n>
(a
1+a2+···+a n)/
√
An=
√
An, so the series
β
b ndiverges. Alsob n/an=
1/
√
An→0.
Section 9.3
In this short section, we present some results that often enable one to handle
series that are conditionally convergent. The easiest and most useful one is the
Alternating Series Test 9.3.2, since alternating series often arise (e.g., from power
series with positive coeﬃcients). In addition, the estimate for the rapidity of
convergence (in Exercise 2) is particularly easy to apply. The tests due to Dirichlet
and Abel are more complicated, but apply to more general series.
Sample Assignment: Exercises 1, 2, 5, 7, 9, 10. (Warning: Exercises 11 and
15(c,f) are rather diﬃcult.)
Partial Solutions:
1. (a) Absolutely convergent. (b) Conditionally convergent.
(c) Divergent. (d) Conditionally convergent.
2. Show by induction thats
2<s4<s6<···<s 5<s3<s1. Hence the limit lies
betweens
nands n+1so that|s−s n|<|s n+1−sn|=zn+1.
3. Letz
2n−1:= 1/nandz 2n:= 0. Or, if it is desired to havez n>0 for alln, take
z
2n−1:= 1/nandz 2n:= 1/n
2
.
4. Let (y
n):=(+1,−1,+1,−1,...).
5. One can use Dirichlet’s Test with (y
n):=(+1,−1,−1,+1,+1,−1,−1,...)to
establish the convergence. Or group the terms in pairs (after the ﬁrst), use
the Alternating Series Test to establish the convergence of the grouped series,
and note that|s
2n−s2n−1|=1/2n→0 so that|s k−s|→0.
6. Ifq>p, thenX:= (1/n
q−p
) is a convergent monotone sequence. Now apply
Abel’s Test withy
n:=an/n
p
.
7. Iff(x):=(lnx)
p
/x
q
, thenf
ﬃ
(x)<0 forxsuﬃciently large. L’Hospital’s Rule
shows that the terms in the alternating series approach 0.
8. (a) Convergent by 9.3.2.
(b) Divergent; use 9.2.1 withy
n:= 1/(n+ 1).
(c) Divergent; the terms do not approach 0.
(d) Divergent; use 9.2.1 withy
n:= 1/n.

Chapter 9 — Infinite Series 73
9. Ift>0, the sequence (e
−nt
) decreases to 0, so Dirichlet’s Test applies.
10. The convergence of
β
(a
n/n) follows from Dirichlet’s Test; the convergence of
β
(s
n/n(n+ 1)) follows by comparison with
β
(1/n
2
). To obtain the equal-
ity, use Abel’s Lemma withx
k:= 1/k, yk:=akandn= 0. Then letm→∞.
11. Dirichlet’s Test does not apply (directly, at least), since the partial sums of
the series generated by (1,−1,−1,1,1,1,...) are not bounded. To estab-
lish the convergence, one can group the terms 1−(1/2+1/3)+(1/4+1/5+
1/6)− ···to get an alternating series. The block consisting ofkterms
ends withn
k:=1+2+···+k=k(k+1)/2, and starts withn k−1+ 1. The
sum of this block ofkterms is greater than the integral of 1/xover the
interval [n
k−1+1,n k+ 1] and less than the integral of 1/xover the interval
[n
k−1,nk]; hence this sum is greater than ln[(n k+1)/(n k−1+ 1)] and less than
ln[n
k/nk−1]. Since it is seen thatn k+1/nk<(nk+1)/(n k−1+ 1) whenk>2,
it follows that the terms in the grouped series are decreasing. Moreover, since
ln(n
k/nk−1)→0, it follows that the terms of the grouped series approach 0;
consequently, the grouped series converges. This means that the subsequence
(s
nk) of the partial sums of the original series converges. But it is readily
seen that ifn
k−1≤n≤n k, thens nlies between the partial sumss nk−1and
s
nk. Hence lim(s n) = lim(s nk), and the series converges.
12. Let|s
n|≤Bfor alln.Ifm>n, then it follows from Abel’s Lemma that
|
β
m
k=n+1
xkyk|≤B[|x m|+|x n+1|+
β
m
k=n+1
|xk−xk+1|]. Sincex n→0 and
β
|x
k−xk+1|is convergent, the dominant term approaches 0, and the Cauchy
Criterion applies.
13. Since (a
n) and (b n) are bounded monotone sequences, they are convergent.
Hence, ifε>0 is given, there existsM(ε) such that ifm>n≥M(ε),
then 0≤a
n+1−am+1<εand 0≤b m+1−bn+1<ε. Sincex k−xk+1=
(a
k−ak+1)+(b k+1−bk), one has
β
m
k=n+1
|xk−xk+1|=(a n+1−am+1)+
(b
m+1−bn+1)<2ε.
14. By Abel’s Lemma,
β
m
k=n+1
ak/k=s m/m−s n/(n+1)+
β
m−1
n+1
sk/k(k+1).
Thus|
β
m
n+1
ak/k|≤M[(1/m)
1−r
+(1/(n+ 1))
1−r
+
β
m−1
n+1
1/k
2−r
]. Since
1−r>0, the ﬁrst two terms approach 0; sincep:= 2−r>1, the series
β
1/k
p
is convergent by 9.2.7(d), so the ﬁnal terms tends to 0.
15. (a) Use Abel’s Test withx
n:= 1/n.
(b) Use the Cauchy Inequality withx
n:=
√
an,yn:= 1/n, to get
β√
an/n≤
(
β
a
n)
1/2
(
β
1/n
2
)
1/2
, establishing convergence.
(c) Let
β
(−1)
k−1
ck,ck>0, be conditionally convergent. Sinceπ/2>3/2,
each intervalI
k:= [(k−1)π+π/4,(k−1)π+3π/4] contains at least one inte-
ger point; we letn
k∈Ikbe the integer nearest (k−1/2)πso that|sinn k|>
1/2. Leta
nk:= (−1)
k−1
ckanda n:= 0 ifn =n kso that
β
a nis convergent.
However, (−1)
k−1
sinn k>1/2 so thatb nk:=anksinn k>ck/2; hence
β
b nis
divergent.

74 Bartle and Sherbert
(d) Leta
n:= [n(lnn)
2
]
−1
, which converges by the Integral Test. However,
b
n:= [
√
nlnn]
−1
, which diverges.
(e) The sequence (n
1/n
) decreases to 1 (see Example 3.1.11(d)); hence Abel’s
Test 9.3.5 applies to give convergence.
(f) If
β
a
nis absolutely convergent, so is
β
b n; otherwise
β
b nmay diverge.
Indeed, (1/k)/(1+1/k)=1/(k+ 1); hence, if the block of terms 1/p,−1/2p,
−1/2pappears in
β
a
n, then the sum of the corresponding block of terms
in
β
b
nis 1/(p+1)−2/(2p+1)=−1/(p+ 1)(2p+ 1). Consequently, if this
block of three terms is repeated 2p+ 1 times in
β
a
n, the sum of the cor-
responding terms in
β
b
nis−1/(n+ 1). Now let (a n) consist of the block
1/1,−1/2,−1/2 repeated 3 times, followed by the block 1/2,−1/4,−1/4
repeated 5 times, followed by the block 1/3,−1/6,−1/6 repeated 7 times,
and so on. Then
β
a
nconverges to 0, but
β
b n=−
β
1/(p+ 1) is divergent.
Section 9.4
The notion of convergence [respectively, uniform convergence] of a series of func-
tions is nothing more than the convergence [resp., uniform convergence] of the
sequence of partial sums of the functions. The importance of the uniform con-
vergence is that it enables one to interchange limit operations (as in Theorems
9.4.2–9.4.4). While the WeierstrassM-Test 9.4.6 is only a suﬃcient condition for
uniform convergence, it is often very useful.
The use of the Ratio Test to determine the radius of convergence will be
familiar to most students, but since the limit of|a
n+1/an|does not always exist,
the Cauchy-Hadamard Theorem 9.4.9 (which always applies) is very important.
It is stressed that the results in 9.4.11 and 9.4.12 arefor power series only.
For general series of functions, one may not be able to integrate or diﬀerentiate
the series term-by-term.
Sample Assignment: Exercises 1(a,c,e), 2, 5, 6(a,c,e), 7, 11, 15, 16, 17.
Partial Solutions:
1. (a) TakeM
n:= 1/n
2
in the WeierstrassM-Test.
(b) Ifa>0, takeM
n:= (1/a
2
)/n
2
to show uniform convergence for|x|≥a.
The series is convergent for allx = 0, but it is not uniformly convergent on
R\{0}, since ifx
n:= 1/n, thenf n(xn)=1.
(c) Since|siny|≤|y|, the series converges for allx. But sincef
n(n
2
)=
sin 1>0, the series is not uniformly convergent onR. However, ifa>0,
the series is uniformly convergent for|x|≤asince then|f
n(x)|≤a/n
2
.
(d) If 0≤x≤1, thenth term does not go to 0, so the series is divergent. If
1<x<∞, the series is convergent, since (x
n
+1)
−1
≤(1/x)
n
. It is uniformly
convergent on [a,∞) fora>1. However, it is not uniformly convergent on
(1,∞); takex
n:= (1 + 1/n)
1/n
.

Chapter 9 — Infinite Series 75
(e) Since 0≤f
n(x)≤x
n
, the series is convergent on [0,1) and uniformly con-
vergent on [0,a] for anya∈(0,1). It is not uniformly convergent on [0,1);
takex
n:= 1−1/n. The series is divergent on [1,∞) since the terms do not
approach 0.
(f) Ifx≥0, the series is alternating and is convergent with|s(x)−s
n(x)|≤
1/(n+ 1). Hence it is uniformly convergent.
2. Since|sinnx|≤1, we can takeM
n:=|a n|.
3. Ifε>0, there existsMsuch that ifn≥M, then|c
nsinnx+···+
c
2nsin 2nx|<εfor allx.Ifx∈[π/6n,5π/12n], then sinkx≥1/2 for
k=n,...,2n, so that (n+1)c
2n<2ε. It follows that 2nc 2n<4εand
(2n+1)c
2n+1<4εforn≥M.
4. Ifρ=∞, then the sequence (|a
n|
1/n
) is not bounded. Hence if|x 0|>0, then
there are inﬁnitely manyk∈Nwith|a
k|>1/|x 0|so that|a kx
k
0
|>1. Thus the
series is not convergent whenx
0 =0.
Ifρ= 0 andx
0 = 0, then since|a n|
1/n
<1/2|x 0|for alln≥n 0, it follows
that|a
nx
n
0
|<1/2
n
forn≥n 0, whence
β
a nx
n
0
is convergent.
5. Suppose thatL:= lim(|a
n|/|an+1|) exists and that 0<L<∞. If follows from
the Ratio Test that
β
a
nx
n
converges for|x|<Land diverges for|x|>L.
Therefore it follows from the Cauchy-Hadamard Theorem thatL=R.
[Alternatively, if 0<ε<L, it can be shown by Induction that there exists
m∈Nsuch that|a
m|(L+ε)
−k
<|am+k|<|a m|(L−ε)
−k
. Hence there exists
A>0,B>0 such thatA(L+ε)
−r
<|ar|<B(L−ε)
−r
forr≥m, whence
A
1/r
/(L+ε)<|a r|
1/r
<B
1/r
/(L−ε) forr≥m. We conclude thatρ=1/L,
soL=R.]
IfL= 0 andε>0 is given, then we have|a
n|<ε|a n+1|forn≥n ε, whence
|a
nx
n
|<|a n+1x
n+1
|for|x|≥εso that the terms do not go to 0 for|x|≥ε
and the series diverges for these values. Sinceε>0 is arbitrary, we have
L=R=0.
IfL=∞, givenM>0, there existsn
Msuch that ifn≥n Mthen|a n+1|<
(1/M)|a
n|. Hence if|x|<M/2, we have|a n+1x
n+1
|≤
1
2
|anx
n
|for alln≥n M
and so the series converges for|x|<M/2. But sinceM>0 is arbitrary, we
deduce thatL=R=∞.
For example, takea
n:=
1
2
forneven anda n:= 2 fornodd. HereLdoes
not exist, butR=1.
6. (a)ρ= lim(1/n)=0 soR=∞.
(b)|a
n/an+1|=(n+1)/(1+1/n)
α
,soR=∞.
(c) lim|a
n/an+1|=1/e,soR=1/e.
(d) lim|a
n/an+1|= 1. Alternatively, since 1/n≤lnn≤n, we have 1/n
1/n
≤
[1/lnn]
1/n
≤n
1/n
,soρ= 1 andR=1.
(e) lim|a
n/an+1|=4, soR=4.
(f) Since lim(n
1/
√
n
)=1,wehaveR=1.

76 Bartle and Sherbert
7. Since 0≤a
n≤1 for alln, buta n= 1 for inﬁnitely manyn, it follows that
ρ= lim sup|a
n|
1/n
= 1, whenceR=1.
8. Note that|a
n|
1/n
≤|nan|
1/n
=n
1/n
|an|
1/n
and use that lim(n
1/n
)=1.
9. By 3.1.11(c) we havep
1/n
→1.
10. By the Uniqueness Theorem 9.4.13,a
n=(−1)
n
anfor alln, so thata n= 0 for
nodd.
11. It follows from Taylor’s Theorem 6.4.1 that if|x|<r, then|R
n(x)|≤
r
n+1
B/(n+ 1)!→0asn→∞.
12. Ifn∈N, there exists a polynomialP
nsuch thatf
(n)
(x)=e
−1/x
2
Pn(1/x)
forx =0.
13. Letg(x):=0 forx≥0 andg(x):=e
−1/x
2
forx<0. Show thatg
(n)
(0)=0
for alln.
14. Ifm∈N, the series reduces to a ﬁnite sum and holds for allx∈R, so we con-
sider the case wheremis not an integer. Ifx∈[0,1) andn>m,n∈N, then by
Taylor’s Theorem there existsc
x∈(0,x) such that 0≤R n(x)≤
⊇
m
n+1
∞
x
m+1
(1 +c x)
n+1−m
≤
⊇
m
n+1
∞
x
n+1
. Use Theorem 3.2.11 to show thatR n(x)→0
asn→∞.
15. Heres
n(x)=(1−x
n+1
)/(1−x).
16. Substitute−yforxin Exercise 15 and integrate fromy=0 toy=xfor
|x|<1, which is justiﬁed by Theorem 9.4.11.
17. If|x|<1, it follows from Exercise 15 that (1 +x
2
)
−1
=
β
∞
n=0
(−1)
n
x
2n
.Ifwe
apply Theorem 9.4.11 and integrate from 0 tox, we get the given expansion
for Arctanx, valid for|x|<1.
18. If|x|<1, it follows from Exercise 14 that (1−x
2
)
1/2
=
β
∞
n=0
⊇
1/2
n
∞
(−1)
n
x
2n
.
Now integrate from 0 toxand evaluate the binomial coeﬃcient.
19. Integratee
−t
2
=
β
∞
n=0
(−1)
n
t
2n
/n! to get

x
0
e
−t
2
dt=
∞
⊂
n=0
(−1)
n
x
2n+1
n!(2n+1)
forx∈R.
20. Apply Exercise 14 and the fact that

π/2
0
(sinx)
2n
dx=
π
2
·
1·3·5···(2n−1)
2·4·6···2n
.

CHAPTER 10
THE GENERALIZED RIEMANN INTEGRAL
This chapter will certainly be new for the students, and it is also likely that it
contains material that will not be familiar to most instructors. However, the
close parallel between Section 10.1 and Sections 7.1–7.3 should make it easier to
absorb the material. Indeed, the only diﬀerence between the generalized Riemann
integral and the ordinary Riemann integral is that slightly diﬀerent orderings are
used for the collection of tagged partitions. It is quite surprising that such a
“slight diﬀerence” in the ordering of the partitions makes such a big diﬀerence in
the resulting classes of integrable functions.
As we have noted the material in the ﬁrst part of Section 10.1 is very similar
to that in Chapter 7. In Section 10.2 we learn that there is no such thing as an
“improper integral”, and that the generalized Riemann integral is not an “abso-
lute” integral. Section 10.3 shows how to extend the integral to functions whose
domain is not bounded; while this procedure seems a bit unnatural, it is quite
simple. (Section 10.3 can be omitted if time is short.)
The ﬁnal Section 10.4 contains some important results; especially the Mono-
tone and Dominated Convergence Theorems. Most treatments of the Lebesgue
integralstartwith the notion of a measurable function, but using our approach it
almost seems to be an afterthought. That is not the case, but just a reﬂection of
the fact that all of the functions we have been dealing with are measurable.
Section 10.1
In order to use the deﬁnition to show that a function is inR
∗
[a, b], we need to
construct a set of gaugesδ
ε. This is done for the speciﬁc functions in Example
10.1.4. (Usually that is rather diﬃcult, except for ordinary Riemann integrable
functions where a constant gauge suﬃces.) In most of the other results in this
section, these gauges are constructed from other gauges; thus a gauge forf+g
is constructed from gauges forfandg. In the Fundamental Theorem 10.1.9, the
gauge forf=F
ﬃ
is constructed using the diﬀerentiability ofF. The Fundamental
Theorems are the highpoint of this section; the later material can be treated more
lightly.
Sample Assignment: Exercises 1, 4, 7(a,c,e), 11, 13, 15, 16, 20.
Partial Solutions:
1. (a) Sincet
i−δ(t i)≤x i−1andx i≤ti+δ(t i), then 0≤x i−xi−1≤2δ(t i).
(b) Apply (a) to each subinterval.
(c) If
˙
Q={([y
j−1,yj],sj)}
m
j=1
satisﬁes
˙
Q ≤δ ∗, thens j−δ(s j)≤s j−δ∗≤
y
j−1andy j≤sj+δ∗≤sj+δ(s j), so thats j∈[yj−1,yj]⊆[s j−δ∗,sj+δ∗]⊆
[s
j−δ(s j),sj+δ(s j)]. Thus
˙
Qisδ-ﬁne.
(d) inf{1/2
k+2
}=0.
77

78 Bartle and Sherbert
2. (a) Iftis a tag for two subintervals, it belongs to both of them, so it is the
right endpoint of one and the left endpoint of the other subinterval.
(b) Consider the tagged partition{([0,1],1),([1,2],1),([2,3],3),([3,4],3)}.
3. (a) If
˙
P={([x
i−1,xi],ti)}
n
i=1
and ift kis a tag for both subintervals [x k−1,xk]
and [x
k,xk+1], we must havet k=xk. We replace these two subintervals by
the subinterval [x
k−1,xk+1] with the tagt k, keeping theδ-ﬁneness property.
Sincef(t
k)(xk−xk−1)+f(t k)(xk+1−xk)=f(t k)(xk+1−xk−1), this consol-
idation of the subintervals does not change the value of the Riemann sums.
A ﬁnite number of such consolidations will result in the desired partition
˙
Q
1.
(b) No. The tagged partition{([0,1],0)([1,2],2)}of [0,2] has the property
that every tag belongs to exactly one subinterval.
(c) Ift
kis the tag for the subinterval [x k−1,xk] and is an endpoint of
this subinterval, we make no change. However, ift
k∈(x k−1,xk), then we
replace [x
k−1,xk] by the two intervals [x k−1,tk] and [t k,xk] both tagged byt k,
keeping theδ-ﬁneness property. Sincef(t
k)(xk−xk−1)=f(t k)(tk−xk−1)+
f(t
k)(xk−tk), this splitting of a subinterval into two subintervals does not
change the value of the Riemann sums.
4. Ifx
k−1≤1≤x kand ift kis the tag for [x k−1,xk], then we cannot have
t
k>1, since thent k−δ(t k)=
1
2
(tk+1)>1. Similarly, we cannot have
t
k<1, since thent k+δ(t k)=
1
2
(tk+1)<1. Therefore we must havet k=1.
If the subintervals [x
k−1,xk] and [x k,xk+1] both have the number 1 as
tag, then 1−.01=1−δ(1)≤x
k−1<xk+1≤1+δ(1)=1+.01 so that
x
k+1−xk−1≤0.02.
5. (a) Letδ(t):=
1
2
min{|t−1|,|t−2|,|t−3|}ift =1,2,3 andδ(t):=1 for
t=1,2,3.
(b) Letδ
2(t) := min{δ(t),δ 1(t)}, whereδis as in part (a).
6. Iff∈R
∗
[a, b] andε>0 is given, then there existsδ εas in Deﬁnition 10.1.1,
and we letγ
ε:=δε.If
˙
Psatisﬁes the stated condition, then
˙
Pisδ ε-ﬁne and
so|S(f;
˙
P)−L|<ε.
Conversely, suppose the stated condition is satisﬁed for some gaugeγ
ε, and
letδ
ε:=
1
2
γε.If
˙
Pisδ ε-ﬁne, then 0<x i−xi−1≤2δ ε(ti)=γ ε(ti), so the
hypothesis implies that|S(f;
˙
P)−L|<ε. Thereforef∈R
∗
[a, b] in the sense
of Deﬁnition 10.1.1.
7. (a)F
1(x):=(2/3)x
3/2
+2x
1/2
,
(b)F
2(x):=(2/3)(1−x)
3/2
−2(1−x)
1/2
,
(c)F
3(x):=(2/3)x
3/2
(lnx−2/3) forx∈(0,1] andF 3(0) := 0,
(d)F
4(x):=2x
1/2
(lnx−2) forx∈(0,1] andF 4(0) := 0,
(e)F
5(x):=−
√
1−x
2
+ Arcsinx.
(f)F
6(x) := Arcsin(x−1).
8. Although the partition
˙
P
0in the proof of 7.1.5 may beδ ε-ﬁne for some gauge
δ
ε, the tagged partition
˙
P zneed not beδ ε-ﬁne, since the valueδ ε(z)maybe

Chapter 10 — The Generalized Riemann Integral 79
much smaller thanδ
ε(xj). For the ordinary Riemann integral, we were only
concerned with the norms
˙
P
0,
˙
P z, which are equal.
9. Iffwere integrable, then

1
0
f≥

1
0
sn=1/2+1/3+···+1/(n+ 1).
10. We enumerate the nonzero rational numbers asr
k=m k/nkand deﬁne
δ
ε(mk/nk):=ε/(n k2
k+1
) andδ ε(x) := 1 otherwise.
11. The functionFis continuous on [a, b], andF
ﬃ
(x)=f(x) forx∈[0,1]\Q.
SinceQis countable, the Fundamental Theorem 10.1.9 applies.
12. The functionMis not continuous on [−2,2], so Theorem 10.1.9 does not
apply. In fact, by Exercise 9 the functionx →1/xis not inR
∗
[0,2] no
matter how we deﬁne it at 0.
13. In fact,L
1is continuous andL
ﬃ
1
(x)=l 1(x) forx = 0, so Theorem 10.1.9
applies.
14. (a) This is possible sinceFis continuous atc
k.
(b) Sincef(c
k) = 0, then we have|F(x i)−F(x i−1)−f(c k)(xi−xi−1)|≤
|F(x
i)−F(c k)|+|F(x i−1)−F(c k)|≤ε/2
k+1
.
(c) The pointc
kcan be the tag for at most two subintervals. The sum of
such terms with tags inEis<ε, and the sum of the terms with tags inI\E
is<ε(b−a).
15. SinceC
ﬃ
1
(x)=(3/2)x
1/2
cos(1/x)+x
−1/2
sin(1/x) forx>0, this function is
inR
∗
[0,1]. Since the ﬁrst term inC
ﬃ
1
has a continuous extension on [0,1], it
is integrable; therefore the second term in also integrable.
16. We haveC
ﬃ
2
(x) = cos(1/x)+(1/x) sin(1/x) forx>0. By the analogue of
Exercise 7.2.12, the ﬁrst term belongs toR[0,1] and therefore toR
∗
[0,1].
17. (a) Takex=ϕ(t):=t
2
+t−2soϕ
ﬃ
(t)=2t+ 1 andE ϕ=∅to get

x=10
x=4
sgnxdx=|10|−|4|=6.
(b) Takex=ϕ(t):=
√
tsox
2
=t,ϕ
ﬃ
(t)=1/(2
√
t) andE ϕ={0}. We get

x=2
x=0
2x
2
(1 +x)
−1
dx=2

2
0
(x−1+(1+x)
−1
)dx= 2(2 + ln 3).
(c) Takex=ϕ(t):=
√
t−1 so thatt=x
2
+1,ϕ
ﬃ
(t)=1/(2
√
t−1) and
E
ϕ={1}. We get

x=2
x=0
2(x
2
+1)
−1
dx= 2 Arctan 2.
(d) Takex=ϕ(t) := Arcsintsot= sinx,ϕ
ﬃ
(t)=(1−x
2
)
−1/2
andE ϕ={1}.
We get

x=π/2
x=0
cos
2
xdx=
1
2

π/2
0
(1 + cos 2x)dx=(
12
x+
1
4
sin 2x)|
π/2
0
=
14
π.
18. Letf(x):=1/
√
xforx∈(0,1] andf(0) := 0 and use Exercise 9.
19. (a) In factf(x):=F
ﬃ
(x) = cos(π/x)+(π/x) sin(π/x) forx>0; we set
f(0) := 0,F
ﬃ
(0) = 0. Thenfand|f|are continuous at every point in (0,1].
It follows as in Exercise 16 thatf∈R
∗
[0,1].
(b) SinceF(a
k)=0and F(b k)=(−1)
k
/k, Theorem 10.1.9 implies that
1/k=|F(b
k)−F(a k)|=|

bk
ak
f|≤

bk
ak
|f|.
(c) If|f|∈R
∗
[0,1], then
β
n
k=1
1/k≤
β
n
k=1

bk
ak
|f|≤

1
0
|f|for alln∈N,
which is a contradiction.

80 Bartle and Sherbert
20. Indeed, sgn(f(x))=(−1)
k
=m(x)on[a k,bk]som(x)·f(x)=|m(x)f(x)|
forx∈[0,1]. Since the restrictions ofmand|m|to every interval [c,1] for
0<c<1 are step functions, they belong toR[c,1]. By Exercise 7.2.11,m
and|m|belong toR[0,1] and

1
0
m=
β
∞
k=1
(−1)
k
/k(2k+ 1) and

1
0
|m|=
β
∞
k=1
1/k(2k+ 1).
21. Indeed,ϕ(x)=Φ
ﬃ
(x)=|cos(π/x)|+(π/x) sin(π/x)·sgn(cos(π/x)) forx ∈E
by Example 6.1.7(c). Evidentlyϕis not bounded near 0. It is seen that
ifx∈[a
k,bk], thenϕ(x)=|cos(π/x)|+(π/x)|sin(π/x)|=|ϕ(x)|so that

bk
ak
|ϕ|=Φ(b k)−Φ(a k)=1/k, from which it follows that|ϕ|/∈R
∗
[0,1].
22. Hereψ(x)=Ψ
ﬃ
(x)=2x|cos(π/x)|+πsin(π/x)·sgn(cos(π/x)) forx/∈
{0}∪E
1by Example 6.1.7(c). Sinceψis bounded, Exercise 7.2.11 applies.
We cannot apply Theorem 7.3.1 to evaluate

b
0
ψsinceEis not ﬁnite, but
Theorem 10.1.9 applies andψ∈R[0,1]. Corollary 7.3.15 implies that|ψ|∈
R[0,1].
23. Ifp≥0, thenmp≤fp≤Mp, wheremandMdenote the inﬁmum and the
supremum offon [a, b], so thatm

b
a
p≤

b
a
fp≤M

b
a
p.If

b
a
p= 0, the
result is trivial; otherwise, the conclusion follows from Bolzano’s Intermediate
Value Theorem 5.3.7.
24. By the Multiplication Theorem 10.1.14,fg∈R
∗
[a, b]. Ifgis increasing,
theng(a)f≤fg≤g(b)fso thatg(a)

b
a
f≤

b
a
fg≤g(b)

b
a
f. Let
K(x):=g(a)

x
a
f+g(b)

b
x
f, so thatKis continuous and takes all values
betweenK(b) andK(a).
Section 10.2
The proof of Hake’s Theorem (which is omitted) is another instance where one
has to construct a set of gauges for the function; here one uses the gauges of the
restrictions of the function to a sequence of intervals [a, γ
n], whereγ n→b.
It is not possible to overestimate the importance of the Lebesgue integral.
Usually this integral is obtained in avery different way.
Sample Assignment: Exercises 1, 2, 5, 6(a,b), 7(a,c,e), 9, 11.
Partial Solutions:
1. Indeed

c
a
f→Aasc→b−if and only if the sequential condition holds.
2. (a) IfG(x):=3x
1/3
forx∈[0,1] then

1
c
g=G(1)−G(c)→G(1)=3.
(b) We have

1
c
(1/x)dx=lnc, which does not have a limit inRasc→0.
3. Here

c
0
(1−x)
−1/2
dx=2−2(1−c)
1/2
→2asc→1−.
4. Since

b
γ
ω= limc→b

c
γ
ω, givenε>0 there existsγ ε≥γsuch that ifγ ε≤
c
1<c2<b, then|

c2
a
f−

c1
a
f|≤

c2
c1
ω<ε. By the Cauchy Criterion, the
limit lim
c→b

c
a
fexists. Now apply Hake’s Theorem.

Chapter 10 — The Generalized Riemann Integral 81
5. Because of continuity,g
1∈R
∗
[c,1] for allc∈(0,1). Ifω(x):=x
−1/2
, then
|g
1(x)|≤ω(x) for allx∈[0,1]. The “left version” of the preceding exercise
implies thatg
1∈R
∗
[0,1] and the above inequality and the Comparison Test
10.2.4 implies thatg
1∈L[0,1].
6. (a,b) Both functions are bounded on [0, 1] (use L’Hospital) and continuous
in (0, 1).
(c) Ifx∈(0,
1
2
] the integrand is dominated by|(ln
1
2
)lnx|.Ifx∈[
1
2
,1) the
integrand is dominated by|(ln
1
2
) ln(1−x)|.
(d) Ifx∈(0,
1
2
] the integrand is dominated by (2/
√
3)|lnx|.Ifx∈[
1
2
,1), the
integrand is bounded and continuous.
7. (a) Convergent, since|f
1(x)|≤1/
√
x.
(b) Divergent, sincef
2(x)≥1/(2x
3/2
) forx∈(0,1].
(c) Divergent, since−f
3(x)≥ln 2/xforx∈(0,
1
2
].
(d) Convergent, since|f
4(x)|≤2|lnx|on (0,
1
2
] and is bounded on [
1
2
,1).
(e) Convergent, since|f
5(x)|≤|lnx|forx∈(0,1].
(f) Divergent, sincef
6(x)≥1/(x−1) forx∈[
1
2
,1).
8. Iff∈R[a, b], thenfis bounded and is inR
∗
[a, b]. Thus the Comparison
Test 10.2.4 applies.
9. Letf(x):=1/
√
xforx∈(0,1] andf(0) := 0.
10. By the Multiplication Theorem 10.1.4, the productfg∈R
∗
[a, b]. Since
|f(x)g(x)|≤B|f(x)|, thenfg∈L[a, b] andfg≤Bf.
11. (a) Letf(x):=(−1)
k
2
k
/kforx∈[c k−1,ck) andf(1) := 0, where thec kare as
in Example 10.2.2(a). Thenf
+
:= max{f,0}equals 2
k
/kon [c k−1,ck) when
kis even and equals 0 elsewhere. Hence

c2k
0
f
+
=1/2+1/4+···+1/2n,so
f
+
/∈R
∗
[0,1].
(b) From the ﬁrst formula in the proof of Theorem 10.2.7, we havef
+
=
max{f,0}=
1
2
(f+|f|). Thus, iff∈L[a, b], then bothf,|f|∈R
∗
[a, b] and
sof
+
belongs toR
∗
[a, b]. Sincef
+
≥0, it belongs toL[a, b].
12. Ifα≤fandα≤g, thenα≤min{f,g}. The second equality in the proof of
Theorem 10.2.7 implies that 0≤|f−g|=f+g−2 min{f,g}≤f+g−2α.
Thereforef+g−2α∈L[a, b] and the Comparison Theorem 10.2.4 implies
thatf+g−2 min{f,g}∈L[a, b], whence min{f,g}∈R
∗
[a, b].
13. (j) Evidently, dist(f,g)=

b
a
|f−g|≥0.
(jj) Iff(x)=g(x) for allx∈[a, b], then dist(f,g)=

b
a
|f−g|=

b
a
0=0.
(jjj) dist(f,g)=

b
a
|f−g|=

b
a
|g−f|= dist(g,f).
(jv) Since|f−h|≤|f−g|+|g−h|, we have dist(f,h)=

b
a
|f−h|≤

b
a
|f−g|+

b
a
|g−h|= dist(f,g) + dist(g,h).
14. Consider the Dirichlet function.

82 Bartle and Sherbert
15. By 10.2.10(iv),f=f±g∓g≤f±g+g, whencef?g≤f±g.
Similarly,g≤g±f∓f≤g±f+f, whenceg?f≤f±g.
Now combine.
16. If (f
n) converges tofinL[a, b], givenε>0 there existsK(ε/2) such that
ifm, n≥K(ε/2) thenf
m−f<ε/2 andf n−f<ε/2. Therefore
f
m−fn≤f m−f+f−f n<ε/2+ε/2=ε. Thuswemaytake
H(ε):=K(ε/2).
17. Indeed,f
n=

1
0
x
n
dx=1/nandf n−θ=1/n.
18. Ifm>n, theng
m−gn≤1/n+1/m→0. One can takeg:= sgn.
19. Sinceh
2n−hn= 1, there is no suchh∈L[0,1].
20. Herek
n=1/nand we can takek=θthe 0-function, or any other function
inL[0,1] with

1
0
|k|=0.
Section 10.3
Although it is important to extend the integral to functions deﬁned on unbounded
intervals, this section can be omitted if time is short.
Sample Assignment: Exercises 1, 3, 5, 7, 13, 15, 17(a,b), 18(a,b).
Partial Solutions:
1. Letb≥max{a,1/δ(∞)}.If
˙
Pis aδ-ﬁne partition of [a, b], show that
˙
Pis a
δ-ﬁne subpartition of [a,∞).
2. The Cauchy Criterion for the existence of lim
γ

γ
a
fis: givenε>0 there
existsK(ε)≥asuch that ifq>p≥K(ε), then|

q
p
f|=|

q
a
f−

p
a
f|<ε.
3. Iff∈L[a,∞), apply the preceding exercise to|f|. Conversely, if

q
p
|f|<ε
forq>p≥K(ε), then|

q
a
f−

p
a
f|≤

q
p
|f|<εso that lim γ

γ
a
fand
lim
γ

γ
a
|f|exist; thereforef,|f|∈R
∗
[a,∞) and sof∈L[a,∞).
4. Iff∈L[a,∞), the existence of lim
γ

γ
a
|f|implies thatVis a bounded set.
Conversely, ifVis bounded, letv:= supV.Ifε>0, there existsKsuch that
v−ε<

K
a
|f|≤v.IfK≤p<q, we have

q
p
|f|<ε, so the preceding
exercise applies.
5. Iff,g∈L[a,∞), thenf,|f|,gand|g|belong toR
∗
[a,∞), so Example
10.3.3(a) implies thatf+gand|f|+|g|belong toR
∗
[a,∞) and

∞
a
(|f|+|g|)=

∞
a
|f|+

∞
a
|g|. Since|f+g|≤|f|+|g|, it follows that

γ
a
|f+g|≤

γ
a
|f|+

γ
a
|g|≤

∞
a
|f|+

∞
a
|g|, whencef+g≤f+g.
6. Indeed,

γ
1
(1/x)dx=lnγ, which does not have limit asγ→∞. Or, use
Exercise 2 and the fact that

2p
p
(1/x)dx=ln2>0 for allp≥1.

Chapter 10 — The Generalized Riemann Integral 83
7. Sincefis continuous on [1,∞), bothf,|f|∈R
∗
[1,γ] for 1≤γ.Ifγ≤
p<q, then|

q
p
f|≤

q
p
|f|≤K

q
p
(1/x
2
)dx≤K(1/p−1/q). Therefore, by
Exercise 2, bothf,|f|belong toR
∗
[1,∞).
8. Ifγ>0, then

γ
0
cosxdx= sinγ, which does not have a limit asγ→∞.
9. (a) We have

γ
0
e
−sx
dx=(1/s)(1−e
−sγ
)→1/s.
(b) LetG(x):=−(1/s)e
−sx
forx∈[0,∞) andG(∞):=0, soGis continuous
on [0,∞) andG(x)→G(∞). By the Fundamental Theorem 10.3.5, we have

∞
0
g=G(∞)−G(0)=1/s.
10. (a) Integrate by parts to get 1/s
2
plus a term that→0asγ→∞.
(b) LetG
1(x):=−(x/s)e
−sx
−(1/s
2
)e
−sx
forx∈[0,∞) andG 1(∞):=0, so
thatG
ﬃ
1
(x)=xe
−sx
andG 1(x)→0asx→∞. The Fundamental Theorem
implies that

∞
0
xe
−sx
dx=G 1(∞)−G 1(0)=1/s
2
.
11. Use Mathematical Induction. The casen= 1 is Exercise 10. Assuming the
formula holds fork∈N, we integrate by parts.
12. (a) Ifx≥e, then (lnx)/x≥1/x. Since

∞
1
(1/x)dxis not convergent, neither
is the given one.
(b) Integrate by parts on [1,γ] and then letγ→∞.
13. (a) Since|sinx|≥1/
√
2>1/2 and 1/x >1/(n+1)πforxin the interval
(nπ+π/4,nπ+3π/4), then|(1/x) sinx|≥1/(2π(n+ 1)) on this interval,
which has lengthπ/2. Therefore

(n+1)π
nπ
|(1/x) sinx|dx≥1/(4(n+ 1)).
(b) Ifγ>(n+1)π, then

γ
0
|D|≥(1/4)(1/1+1/2+···+1/(n+ 1)).
14. The integrand is bounded, so is inR
∗
[0,γ]. Integrating by parts, we get

q
p
x
−1/2
sinxdx=−x
−1/2
cos

q
p
−(1/2)

q
p
x
−3/2
cosxdx. Since|cosx|≤1,
we have|

q
p
x
−1/2
sinxdx|≤q
−1/2
+p
−1/2
+(1/2)

q
p
x
−3/2
dxwhich is
≤(5/4)(q
−1/2
+p
−1/2
)→0asp→∞.
15. Letu=ϕ(x)=x
2
so that

γ
0
sin(x
2
)dx=(1/2)

γ
2
0
u
−1/2
sinudu.Now
apply Exercise 14.
16. (a) Convergent. Since lnx∈R
∗
[0,γ] the given integrand is inR
∗
[0,γ]. Since
(lnx)/
√
x→0asx→∞, then|(lnx)/(x
2
+1)|≤K/x
3/2
forxsuﬃciently
large.
(b) Divergent. As in (a), the integrand is inR
∗
[0,1]. Since 4x
2
>x
2
+1
forx≥1, then (lnx)/
√
x
2
+1>(lnx)/2x≥1/2xforx≥e, so that the
integrand is not inR
∗
[e,∞].
(c) Divergent. Ifx∈[0,1], then 2>x+ 1 so that 1/x(x+1)>1/2x.Thus
the restriction of the integrand is not inR
∗
[0,1], so the integrand is not in
R
∗
[0,∞].
(d) Convergent. The integrand is dominated by 1/x
2
on [0,∞].
(e) Divergent.x/
3
√
1+x
3
→1asx→∞, so that 1/2<x/
3
√
1+x
3
forx
suﬃciently large, so 1/(2x)<1/
3
√
1+x
3
forxsuﬃciently large.

84 Bartle and Sherbert
(f) Convergent. Since 0≤Arctanx≤π/2 forx≥0 so the integrand is
dominated by 1/(x
3/2
+1)<1/x
3/2
forx≥1.
17. (a) Iff
1(x) := sinx, thenf 1/∈R
∗
[0,∞). By Exercise 14, iff 2(x):=x
−1/2
sinx,
thenf
2∈R
∗
[0,∞) andϕ 2(x):=1/
√
xis bounded and decreasing on [1,∞).
(b) Iff(x) := sin(x
2
), then Exercise 15 implies thatf∈R
∗
[0,∞). Here
ϕ(x):=x/(x+ 1) is bounded and increasing on [0,∞).
(c) Takef(x):=x
−1/2
sinx∈R
∗
[0,∞) by Exercise 14 andϕ(x):=(x+1)/x
so thatϕis bounded and decreasing on [0,∞).
(d) Takeϕ(x) := Arctanx,soϕis bounded and increasing on [0,∞), while
f(x):=1/(x
3/2
+1)<x
−3/2
forx≥1.
18. (a)f(x) := sinxis continuous so is inR
∗
[0,γ]. AlsoF(x):=

x
0
sintdt=
1−cosxis bounded by 2 on [0,∞) andϕ(x):=1/xdecreases to 0.
(b) Takeϕ(x):=1/lnxsoϕdecreases monotonely to 0.
(c)F(x):=

x
0
costdt= sinxis bounded by 1 on [0,∞) andϕ(x):=x
−1/2
decreases monotonely to 0.
(d)ϕ(x):=x/(x+ 1) increases to 1 (not 0).
19. Letu=ϕ(x):=x
2
so that

γ
0
x
1/2
sin(x
2
)dx=(1/2)

γ
2
0
u
−1/4
sinudu.By
the Chartier-Dirichlet Test, this integral converges and Hake’s Theorem
applies.
20. (a) Ifγ>0, then

γ
0
e
−x
dx=1−e
−γ
→1soe
−x
∈R
∗
[0,∞). Similarly
e
−|x|
=e
x
∈R
∗
(−∞,0].
(b)|x−2|/e
−x/2
→0asx→∞,so|x−2|≤e
−x/2
forxsuﬃciently large,
so the integrand is dominated bye
−x/2
forxlarge. Therefore the integrand
is inR
∗
[0,∞) and similarly on (−∞,0].
(c) We have 0≤e
−x
2
≤e
−x
for|x|≥1, soe
−x
2
∈R
∗
[0,∞). Similarly on
(−∞,0].
(d) The integrand approaches 1 asx→0. Sincee
x
/(e
x
−e
−x
)→1as
x→∞, we have 2x/(e
x
−e
−x
)≤4xe
−x
forxsuﬃciently large. Therefore
the integrand is inR
∗
[0,∞). Similarly on (−∞,0].
Section 10.4
This section contains some very important results.
Sample Assignment: Exercises 1, 3(a,c,e), 5, 6, 9, 11, 14.
Partial Solutions:
1. (a) Converges to 0 atx= 0, to 1 on (0, 1]. Not uniform. Bounded by 1.
Increasing. Limit = 1.
(b) Converges to 0 on [0, 1), to
1
2
atx= 1, to 1 on (1,2]. Not uniform.
Bounded by 1. Not monotone (although decreasing on [0,1] and increasing
on [1,2]). Limit = 1.

Chapter 10 — The Generalized Riemann Integral 85
(c) Converges to 1 on [0, 1), to
1
2
atx= 1. Not uniform. Bounded by 1.
Increasing. Limit = 1.
(d) Converges to 1 on [0, 1), converges to
1
2
atx= 1, to 0 on (1, 2]. Not
uniform. Bounded by 1. Not monotone (although increasing on [0, 1] and
decreasing on (1, 2]). Limit = 1.
2. (a) Converges to
√
xon [0, 1]. Uniform. Bounded by 1. Increasing. Limit =
2/3.
(b) Deﬁne to equal 0 atx= 0, converges to 1/
√
xon (0, 1), to
1
2
atx=1.
Not uniform. Not bounded. Dominated by 1
√
x. Increasing. Limit = 2.
(c) Converges to
1
2
atx= 1, to 0 on (1, 2]. Not uniform. Bounded by 1.
Decreasing. Limit = 0.
(d) Deﬁne to equal 0 atx= 0, converges to 1/2
√
xon (0, 1), to 1 atx=1.
Not uniform. Not bounded. Dominated by 1/2
√
x. Decreasing. Limit = 1.
3. (a) Converges to 1 atx= 0, to 0 on (0,1]. Not uniform. Bounded by 1.
Decreasing. Limit = 0.
(b) Deﬁne to be 0 atx= 0. The functions do not have (a ﬁnite) integral.
Converges to 0. Not uniform. Not bounded. Decreasing. Integral of limit = 0.
(c) Converges to 0. Not uniform. Bounded by 1/e. Not monotone. Limit = 0.
(d) Converges to 0. Not uniform. Not bounded. Not monotone. Limit =

∞
0
ye
−y
dy=1.
(e) Converges to 0. Not uniform. Bounded by 1/
√
2e. Not monotone.
Limit = 0.
(f) Converges to 0. Not uniform. Not bounded. Not monotone. Not
dominated. Limit =
1
2

∞
0
e
−y
dy=
12
.
4. (a) Sincef
k(x)→0 forx∈[0,1) and|f k(x)|≤1, the Dominated Convergence
Theorem applies.
(b)f
k(x)→0 forx∈[0,1), but (f k(1)) is not bounded. No obvious
dominating function. Integrate by parts and use (a). The result shows that
the Dominated Convergence Theorem does not apply.
5. Note thatf
kis a step function and

2
0
fk=k(1/k)=1. Ifx∈(0,2], there
existsk
xsuch that 2/k < xfork≥k x; thereforef k(x)→0.
6. Suppose that (f
k(c)) converges for somec∈[a, b]. By the Fundamental
Theorem, we havef
k(x)−f k(c)=

x
c
f
ﬃ
k
. By the Dominated Convergence
Theorem,

x
c
f
ﬃ
k
→

x
c
g, whence (f k(x)) converges for allx∈[a, b]. Note that
iff
k(x):=(−1)
k
, then (f k(x)) does not converge for anyx∈[a, b].
7. Indeed,g(x):=sup{f
k(x):k∈N}equals 1/kon (k−1,k], so that

n
0
g=
1+
1
2
+···+
1
n
. Henceg/∈R
∗
[0,∞).
8. Indeed,

∞
0
e
−tx
dx=(−1/t)e
−tx

x=∞
x=0
=1/t. If we integrate by parts, then
we get

∞
0
xe
−tx
dx=(−x/t−1/t
2
)e
−tx

x=∞
x=0
=1/t
2
.
9. Indeed,

∞
0
e
−tx
sinxdx=−[e
−tx
(tsinx+ cosx)/(t
2
+ 1)]

x=∞
x=0
=1/(t
2
+ 1).

86 Bartle and Sherbert
10. (a) Ifa>0, then|(e
−tx
sinx)/x|≤e
−ax
fort∈J a:= (a,∞). Ift k∈Jaand
t
k→t0∈Ja, then the argument in 10.4.6(d) shows thatEis continuous att 0.
Also, ift
k≥1, then|(e
−tkx
sinx)/x|≤e
−x
and the Dominated Convergence
Theorem implies thatE(t
k)→0.ThusE(t)→0ast→∞.
(b) It follows as 10.4.6(e) thatE
ﬃ
(t0)=−

∞
0
e
−t0x
sinxdx=−1/(t
2
0
+ 1).
(c) By 10.1.9,E(s)−E(t)=

s
t
E
ﬃ
(t)dt=−

s
t
(t
2
+1)
−1
dt= Arctant−
Arctansfors, t >0. ButE(s)→0 and Arctans→π/2ass→∞.
(d) We do not know thatEis continuous ast→0+.
11. (a) Note thate
−t
2
(x
2
+1)
≤1 fort≥0 and thate
−t
2
(x
2
+1)
→0ast→∞
for allx≥0. Thus the Dominated Convergence Theorem can be applied to
sequences to give the continuity ofGand the fact thatG(t)→0ast→∞.
(b) The partial derivative equals−2te
−t
2
e
−t
2
x
2
, which is bounded by 2 for
t≥0,x∈[0,1]. An argument as in 10.4.6(e) gives the formula forG
ﬃ
(t).
(c) Indeed,F
ﬃ
(t)=2e
−t
2
t
0
e
−x
2
dxfort≥0, soF
ﬃ
(t)=−G
ﬃ
(t) fort≥0.
(d) Since lim
t→∞F(t)=
1
4
π,wehave

∞
0
e
−x
2
dx=
1
2
√
π.
12. Fixx∈I. As in 10.4.6(e), ift, t
0∈[a, b], there existst xbetweent, t 0
such thatf(t, x)−f(t 0,x)=(t−t 0)
∂f
∂t
(tx,x). Thereforeα(x)≤[f(t, x)−
f(t
0,x)]/(t−t 0)≤ω(x) whent =t 0. Now argue as before and use the
Dominated Convergence Theorem 10.4.5.
13. (a) If (s
k) is a sequence of step functions converging tofa.e., and (t k)isa
sequence of step functions converging toga.e., then it follows from Theorem
10.4.9(a) and Exercise 2.2.18 that (max{s
k,tk}) is a sequence of step functions
that converges to max{f,g}a.e. Similarly, for min{f,g}.
(b) By part (a), the functions max{f,g}, max{g,h}and max{h, f}are mea-
surable. Now apply Exercises 2.2.18 and 2.2.19.
14. (a) Sincef
k∈M[a, b] is bounded, it belongs toR
∗
[a, b]. The Dominated
Convergence Theorem implies thatf∈R
∗
[a, b]. The Measurability Theorem
10.4.11 now implies thatf∈M[a, b].
(b) Sincet →Arctantis continuous, Theorem 10.4.9(b) implies thatf
k:=
Arctan◦g
k∈M[a, b]. Further,|f k(x)|≤
1
2
πforx∈[a, b], so (f k), is also a
bounded sequence inM[a, b].
(c) Ifg
k→ga.e., from the continuity of Arctan, it follows thatf k→fa.e.
Part (a) implies thatf∈M[a, b] and Theorem 10.4.9(b) applied toϕ= tan
implies thatg= tan◦f∈M[a, b].
15. (a) Since1
Eis bounded, it is inR
∗
[a, b] if and only if it is inM[a, b].
(b) Indeed,1
∅is the 0-function, and ifJis any subinterval of [a, b], then1 J
is a step function.
(c) This follows from the fact that1
E
ﬀ=1−1 E.
(d) We havex∈E∪Fif and only ifx∈Eorx∈F.Thus
1
E∪F(x)=1⇐⇒1 E(x)=1or1 F(x)=1⇐⇒max{1 E(x),1 F(x)}=1.

Chapter 10 — The Generalized Riemann Integral 87
Similarly,x∈E∩Fif and only ifx∈E and x∈F.Thus
1
E∩F(x)=1⇐⇒1 E(x)=1and1 F(x)=1⇐⇒min{1 E(x),1 F(x)}=1.
Further,E\F=E∩F
ﬃ
.
(e) If (E
k) is an increasing sequence inM[a, b], then (1 Ek) is an increasing
sequence inM[a, b]. Moreover,1
E(x) = limk1Ek(x), and we can apply
Theorem 10.4.9(c). Similarly, (1
Fk) is a decreasing sequence inM[a, b] and
1
F(x) = limk1Fk(x).
(f) LetA
n:=
ﬀ
n
k=1
Ek, so that (A n) is an increasing sequence inM[a, b] with
ﬀ
∞
n=1
An=E, so (e) applies. Similarly, ifB n:=
ﬃ
n
k=1
Fk, then (B n)isa
decreasing sequence inM[a, b] with
ﬃ
∞
n=1
Bn=F.
16. (a)m(∅)=

b
a
0 = 0 and 0≤1 E≤1 implies 0≤m(E)=

b
a
1E≤

b
a
1=
b−a.
(b) Since1
[c,d]is a step function, thenm([c, d]) =d−c. The other charac-
teristic functions are a.e. to1
[c,d], so have the same integral.
(c) Since1
E
ﬀ=1−1 E,wehavem(E
ﬃ
)=

b
a
(1−1 E)=(b−a)−m(E).
(d) Note that1
E∪F+1E∩F=1E+1F. Therefore,m(E∪F)+m(E∩F)=

b
a
(1E∪F+1E∩F)=

b
a
(1E+1F)=m(E)+m(F).
(e) IfE∩F=∅, then (d) and (a) imply thatm(E∪F)+0=m(E)+m(F).
(f)If(E
k) is increasing inM[a, b]toE, then (1 Ek) is increasing inM[a, b]to
1
E. The Monotone Convergence Theorem 10.4.4 implies that1 E∈M[a, b]
and thatm(E
k)=

b
a
1Ek→

b
a
1E=m(E).
(g) If (C
k) is pairwise disjoint andE n:=
ﬀ
n
k=1
Ckforn∈N, then, by
Induction in part (e), we havem(E
n)=m(C 1)+···+m(C n). But, since
ﬀ
∞
k=1
Ck=
ﬀ
∞
n=1
Enand (E n) is increasing, (f) implies that
m
≈
∞

k=1
Ck

= lim
n
m(E n) = lim
n
n
⊂
k=1
m(C k)=
∞
⊂
n=1
m(C k).

CHAPTER 11
A GLIMPSE INTO TOPOLOGY
We present in this chapter an introduction into the subject of topology. In the ﬁrst
edition of this book, most of the ideas presented here were discussed as the notions
naturally arose. However, our experience in teaching from that edition was that
some of the students were confused by ideas that they felt were very abstract and
diﬃcult. Consequently, in later editions, we have dealt only with open and closed
intervalsin Chapters 1 through 10, even though some of the results that were
established held for general open and closed (or at least compact) subsets ofR.
Some instructors may wish to blend part of this material into their presenta-
tion of the earlier material. Others may decide to omit the entire chapter, or to
assign it only to the better students as a unifying “special project”.
In the ﬁnal section, we give the deﬁnitions of a metric function and a metric
space. They are very important for further developments in analysis as well as in
the ﬁeld of topology, and we feel that they are quite natural ideas. This section
will serve as a springboard to students who continue their study of analysis beyond
this course.
Section 11.1
Here the notions of open and closed subsets ofRare introduced and such sets are
characterized. The ﬁnal topic of this section is the Cantor setF, which should
expand the imagination of the students.
Sample Assignment: Exercises 1, 2, 4, 9, 10, 13, 18, 23.
Partial Solutions:
1. If|x−u|<inf{x,1−x}, thenu<x+(1−x)=1 andu>x−x= 0, so that
0<u<1.
2. Ifx∈(a,∞), then takeε
x:=x−a. The complement of [b,∞) is the open
set (−∞,b).
3. Suppose thatG
1, ...,Gk,Gk+1are open sets and thatG 1∪ ··· ∪G kis open.
It then follows from the fact that the union of two open sets is open that
G
1∪ ··· ∪G k∪Gk+1=(G 1∪ ··· ∪G k)∪G k+1is open.
4. Ifx∈(0,1], thenx∈(0,1+1/n) for alln∈N. Also, ifx>1, thenx−1>0
so there existsn
x∈Nsuch thatx−1>1/n x, whencex/∈(0,1+1/n x).
5. The complement ofNis the union (−∞,1)∪(1,2)∪···of open intervals.
6. The sequence (1/n) belongs toAand converges to 0/∈A,soAis not closed,
by 11.1.7. Alternatively, use 11.1.8 and the fact that 0 is a cluster point ofA.
7. Corollary 2.4.9 of the Density Theorem implies that every neighborhood of a
pointxinQcontains a point not inQ. HenceQis not an open set.
88

Chapter 11 — A Glimpse into Topology 89
8. IfFis a closed set, its complementC(F)isopenandG\F=G∩C(F).
9. This is a rephrasing of Deﬁnition 11.1.2.
10. Note thatxis a boundary point ofA⇐⇒every neighborhoodVofx
contains points inAand points inC(A)⇐⇒xis a boundary point ofC(A).
11. Note that ifA⊆Randx∈R, then precisely one of the following statements
is true: (i)xis an interior point ofA, (ii)xis a boundary point ofA, and (iii)
xis an interior point ofC(A). Hence, ifAis open, then it does not contain
any boundary points (since it contains only interior points) ofA. Conversely,
ifAdoes not contain any boundary points ofA, then all of its points are
interior points ofA.
12. LetFbe closed and letxbe a boundary point ofF.Ifx/∈F, thenx∈C(F).
SinceC(F) is an open set, there exists a neighborhoodVofxsuch that
V⊆C(F), contradicting the hypothesis thatxis a boundary point ofF.
Conversely, ifFcontains all of its boundary points and ify/∈F, thenyis
not a boundary point ofF, so there exists a neighborhoodVofysuch that
V⊆C(F). This implies thatC(F) is open, so thatFis closed.
(Alternative proof.) The setsFandC(F) have the same boundary points.
ThereforeFcontains all of its boundary points⇐⇒ C(F) does not contain
any of its boundary points⇐⇒ C(F)isopen.
13. SinceA
◦
is the union of open sets, it is open (by 11.1.4(a)). IfGis an open set
withG⊆A, thenG⊆A
◦
(by its deﬁnition). Alsox∈A
◦
⇐⇒xbelongs
to an open setV⊆A⇐⇒xis an interior point ofA.
14. SinceA
◦
is the union of subsets ofA, we haveA
◦
⊆A. It follows that
(A
◦
)
◦
⊆A
◦
. SinceA
◦
is an open subset ofA
◦
and (A
◦
)
◦
is the union of
all open sets contains inA
◦
, thenA
◦
⊆(A
◦
)
◦
. ThereforeA
◦
=(A
◦
)
◦
SinceA
◦
is an open set inAandB
◦
is an open set inB, thenA
◦
∩B
◦
is an open set inA∩B, whenceA
◦
∩B
◦
⊆(A∩B)
◦
. And sinceA
◦
∩B
◦
is an open set inA, then (A∩B)
◦
⊆A
◦
; similarly (A∩B)
◦
⊆B
◦
, so that
(A∩B)
◦
⊆A
◦
∩B
◦
. Therefore (A∩B)
◦
=A
◦
∩B
◦
.
IfA:=QandB:=R\Q, thenA
◦
=B
◦
=∅, whileA∪B=R, whence
(A∪B)
◦
=R.
15. SinceA
−
is the intersection of all closed sets containingA, then by 11.1.5(a)
it is a closed set containingA. SinceC(A
−
) is open, thenz∈C(A
−
)⇐⇒z
has a neighborhoodV
ε(z)inC(A
−
)⇐⇒zis neither an interior point nor a
boundary point ofA.
16. For any setB, sinceB
−
is a closed set containingB, thenB⊆B
−
.Ifwe
takeB=A
−
, we getA
−
⊆(A
−
)
−
. SinceA
−
is a closed set containingA
−
,
we have (A
−
)
−
⊆A
−
. Therefore (A
−
)
−
=A
−
.
Since (A∪B)
−
is a closed set containingA, thenA
−
⊆(A∪B)
−
. Similarly
B
−
⊆(A∪B)
−
, so we conclude thatA
−
∪B
−
⊆(A∪B)
−
. ConverselyA⊆A
−
andB⊆B
−
, and sinceA
−
andB
−
are closed, it follows from 11.1.5(b) that
A
−
∪B
−
is closed; hence (A∪B)
−
⊆A
−
∪B
−
.

90 Bartle and Sherbert
IfA:=Q, thenA
−
=R;ifB:=R\Q, thenB
−
=R. ThereforeA∩B=∅
whileA
−
=B
−
=A
−
∪B
−
=R.
17. TakeA=Q.
18. Eitheru:= supFbelongs toF,oruis a cluster point ofF.IfFis closed,
then 11.1.8 implies that any cluster point ofFbelongs toF.
19. IfG =∅is open andx∈G, then there existsε>0 such thatV
ε(x)⊆G,
whence it follows thata:=x−εis inA
x.
20. Ifa
x∈G, then sinceGis open, there existsε>0 with (a x−ε, ax+ε)⊆G.
This contradicts the deﬁnition ofa
x.
21. Ifa
x<y<xthen sincea x:= infA xthere existsa
ﬃ
∈Axsuch thata x<a
ﬃ
≤y.
Therefore (y,x]⊆(a
ﬃ
,x]⊆Gandy∈G.
22. Ifb
x =by, then either (i)b x<byor (ii)b y<bx. In case (i), thenb x∈Iy=
(a
y,by)⊆G, contrary tob x/∈G. In case (ii), thenb y∈Ix=(ax,bx)⊆G,
contrary tob
y/∈G.
23. Ifx∈Fandn∈N, the intervalI
ninFncontainingxhas length 1/3
n
. Let
y
nbe an endpoint ofI nwithy n =x. Then 0<|y n−x|≤1/3
n
. Sincey nis
an endpoint ofI
n, it also belongs toF. Consequentlyx= lim(y n) is a cluster
point ofF.
24. Ifx∈Fandn∈N, the intervalI
ninFncontainingxhas length 1/3
n
. Letz n
be the midpoint ofI n, so that 0<|z n−x|≤1/3
n
. Sincez ndoes not belong to
F
n+1, it follows thatz n∈C(F). Consequentlyx= lim(z n) is a cluster point
ofC(F).
Section 11.2
Most students will ﬁnd the notion of compactness to be diﬃcult, especially when
they learn that they must be prepared to considereveryopen cover of the set.
They also ﬁnd the Heine-Borel Theorem 11.2.5 a bit disappointing, since compact
sets inRturn out to be of a very easily described nature. But this is exactly the
reason why the Heine-Borel Theorem is important: it makes the determination of
a compact set inRa relatively simple matter. Students need to be told that the
situation is diﬀerent in more complicated topological spaces; unfortunately, they
will have to take that fact on faith until a later course.
While the sequential characterizations of compact sets are somewhat special,
they are easier to grasp than the covering aspects.
Sample Assignment: Exercises 1, 3, 4, 5, 6, 9, 11.
Partial Solutions:
1. LetG
n:= (1 + 1/n,3) forn∈N.
2. LetG
n:= (n−1/2,n+1/2) forn∈N.

Chapter 11 — A Glimpse into Topology 91
3. LetG
n:= (1/2n,2) forn∈N.
4. IfGis an open cover ofF, thenG∪{C(F)}is an open cover ofK.
5. IfG
1is an open cover ofK 1andG 2is an open cover ofK 2, thenG 1∪G2is
an open cover ofK
1∪K2.
6. LetKbe a bounded inﬁnite subset ofR; we want to show thatKhas a cluster
point. If not, then it follows from Theorem 11.1.8 thatKmust be closed.
SinceKis bounded, it follows from the Heine-Borel Theorem 11.2.5 thatK
is compact. Ifk∈Kis arbitrary, then sincek∈Kis not a cluster point ofK,
we conclude that there exists an open neighborhoodJ
kofkthat contains no
point ofK\{k}. But since{J
k:k∈K}is an open cover ofK, it follows that
there exists a ﬁnite number of pointsk
1, ...,knsuch that{J ki
:i=1, ...,n}
coversK. But this implies thatKis a ﬁnite set.
7. LetK
n:= [0,n] forn∈N.
8. If{K
α}is a collection of compact subsets ofR, it follows from the Heine-
Borel Theorem 11.2.5 that each setK
αis closed and bounded. Hence, from
11.1.5(a) the setK
0:=
ﬃ
K αis also closed. SinceK 0is also bounded (since
it is a subset of a bounded set), it follows from the Heine-Borel Theorem that
K
0is compact, as asserted.
9. For eachn∈N, letx
n∈Kn. Since the set{x n}⊆K 1, we infer that the
sequence (x
n) is a bounded sequence. By the Bolzano-Weierstrass Theorem,
(x
n) has a subsequence (x mr) that converges to a pointx 0. Sincex mr∈Kn
for allr≥n, it follows thatx 0= lim(x mr) belongs to
ﬃ
K n.
10. SinceK =∅is bounded, it follows that infKexists inR.IfK
n:={k∈K:
k≤(infK)+1/n}, thenK
nis closed and bounded, hence compact. By the
preceding exercise
ﬃ
K
n=∅, but ifx 0∈
ﬃ
K n, thenx 0∈Kand it is readily
seen thatx
0= infK. [Alternatively, use Theorem 11.2.6.]
11. Forn∈N, letx
n∈Kbe such that|c−x n|≤inf{|c−x|:x∈K}+1/n.Now
apply Theorem 11.2.6.
12. LetK⊆Rbe compact and letc∈R.Ifn∈N, there existsx
n∈Ksuch that
sup{|c−x|:x∈K}−1/n <|c−x
n|. It follows from the Bolzano-Weierstrass
Theorem that there exists a subsequence (x
nk) that converges to a pointb,
which also belongs to the compact setK. Moreover, we have
|c−b|= lim|c−x
nk|≥sup{|c−x|:x∈K}.
But sinceb∈k, it also follows that|c−b|≤sup{|c−x|:x∈K}.
13. The family{V
δx
(x):x∈K}forms an open cover of the compact set [a, b].
Therefore it can be replaced by a ﬁnite subcover, say{V
x1, ...,Vxn}.Ifb jis
a bound forfonV
xj, then sup{b 1, ...,bn}is a bound forfon [a, b].
14. SupposeK
1andK 2are disjoint compact sets and assume that inf{|x−y|:
x∈K
1,y∈K 2}= 0. Then there exist sequences (x n)inK 1and (y n)inK 2

92 Bartle and Sherbert
such that|x
n−yn|<1/nforn∈N. Let (x
ﬃ
k
) be a subsequence of (x n) that
converges to a pointx
0∈K1, and let (y
ﬃ
k
) be the corresponding subsequence
of (y
n). Then (y
ﬃ
k
) has a subsequence (y
ﬃﬃ
k
) that converges to a pointy 0∈K2.
If (x
ﬃﬃ
k
) is the corresponding subsequence of (x
ﬃ
k
), then we conclude that
|x
0−y0|= lim|x
ﬃﬃ
k
−y
ﬃﬃ
k
|= 0, from which it follows thatx 0=y0, so thatK 1
andK 2are not disjoint, contrary to the hypothesis.
15. LetF
1:={n:n∈N}andF 2:={n+1/n:n∈N,n≥2}.
Section 11.3
The relationship between continuous functions and open sets is very important and
the interplay between continuous functions and compact sets is further clariﬁed
here. Students who go on to more advanced courses in topology should ﬁnd this
to be a very useful introduction.
Sample Assignment: Exercises 1, 2, 4, 5, 6, 9.
Partial Solutions:
1. (a) Ifa<b≤0, thenf
−1
(I)=∅.Ifa<0<b, thenf
−1
(I)=(−
√
b,
√
b). If
0≤a<b, thenf
−1
(I)=(−
√
b,−
√
a)∪(
√
a,
√
b).
(b) IfI:= (a, b) wherea<0<b, thenf(I)=[0,c), wherec:= sup{a
2
,b
2
}.
2. (a)fmaps the interval (−1,1) to (1/2,1]. (b)fmaps [0,∞)to(0,1].
3.f
−1
(G)=f
−1
([0,ε))=[1,1+ε
2
))=(0,1+ε
2
)∩I.
4. LetG:= (1/2,3/2). LetF:= [−1/2,1/2].
5.f
−1
((−∞,α)) is the inverse image of the open set (−∞,α).
6. The set{x∈R:f(x)≤α}=f
−1
((−∞,α]) and (−∞,α] is closed inR.
7. Ifx
n∈Ris such thatf(x n)=kfor alln∈Nand ifx= lim(x n), then
f(x) = lim(f(x
n)) =k. Alternatively,{x∈R:f(x)=k}={x∈R:f(x)≤k}∩
{x∈R:f(x)≥k}.
8. Letfbe the Dirichlet Discontinuous Function.
9. First note that ifA⊆Randx∈R, then we havex∈f
−1
(R\A)⇐⇒f(x)∈
R\A⇐⇒f(x)/∈A⇐⇒x/∈f
−1
(A)⇐⇒x∈R\f
−1
(A); therefore,
f
−1
(R\A)=R\f
−1
(A). Now use the fact a setF⊆Ris closed if and only
ifR\Fis open, together with Corollary 11.3.3.
10. If (x
n) is a sequence inIsuch thatf(x n)=g(x n) for allnand ifx n→x0, then
it follows from the fact thatI=[a, b] is closed thatx
0∈I. Moreover, sincef
andgare continuous atx
0,thenf(x 0) = limf(x n) = limg(x n)=g(x 0).
Section 11.4
We have merely introduced the notion of a metric space, but we think it may be
useful and stimulating for students to see that a great deal of what has been done

Chapter 11 — A Glimpse into Topology 93
can be extended to much wider generality. Some instructors may wish to use this
brief section as a springboard for further discussion; other may decide to omit it
completely.
Sample Assignment: Exercises 1, 3, 4, 7, 9, 10.
Partial Solutions:
1. IfP
1:=(x 1,y1),P2:= (x 2,y2),P3:= (x 3,y3), then
d
1(P1,P2)≤(|x 1−x3|+|x 3−x2|)+(|y 1−y3|+|y 3−y2|)
=d
1(P1,P3)+d 1(P3,P2).
Thusd
1satisﬁes the Triangle Inequality.
To see thatd
∞satisﬁes the Triangle Inequality, note that|x 1−x3|≤
d
∞(P1,P3) and|y 1−y3|≤d ∞(P1,P3), and also that|x 3−x2|≤d ∞(P3,P2)
and|y
3−y2|≤d∞(P3,P2). Therefore, we have|x 1−x2|≤|x 1−x3|+|x3−x2|≤
d
∞(P1,P3)+d ∞(P3,P2) and|y 1−y2|≤|y 1−y3|+|y 3−y2|≤d∞(P1,P3)+
d
∞(P3,P2), whence it follows thatd ∞(P1,P2) = sup{|x 1−x2|,|y1−y2|} ≤
d
∞(P1,P3)+d ∞(P3,P2).
2. Since|f(x)−g(x)|≤|f(x)−h(x)|+|h(x)−g(x)|≤d
∞(f,h)+d ∞(h, g) for all
x∈[0,1], it follows thatd
∞(f,g)≤d ∞(f,h)+d ∞(h, g) andd ∞satisﬁes the
Triangle Inequality.
We also haved
1(f,g)=

1
0
|f−g|≤

1
0
{|f−h|+|h−g|}=

1
0
|f−h|+

1
0
|h−g|=d 1(f,h)+d 1(h, g).
3. We haves =tif and only ifd(s, t)=1. Ifs =t, the value ofd(s, u)+d(u, t)
is either 1 or 2 depending on whetheruequalssort, or neither.
4. Sinced
∞(Pn,P) = sup{|x n−x|,|y n−y|},ifd ∞(Pn,P)→0 then it follows
that both|x
n−x|→0 and|y n−y|→0, whencex n→xandy n→y. Con-
versely, ifx
n→xandy n→y, then|x n−x|→0 and|y n−y|→0, whence
d
∞(Pn,P)→0.
5. If (x
n),(yn) converge tox,y, respectively, thend(P n,P)=|x n−x|+
|y
n−y|→0, so that (P n) converges toP. Conversely, since|x n−x|≤d(P n,P),
ifd(P
n,P)→0, then lim(x n)=x, and similarly for (y n).
6. If a sequence (x
n)inSconverges toxrelative to the discrete metricd,
thend(x
n,x)→0 which implies thatx n=xfor all suﬃciently largen. The
converse is trivial.
7. Show that a set consisting of a single point is open. Then it follows that
every set is an open set, so that every set is also a closed set.
8. (a){(x, y):|x|+|y|≤1}is the square with vertices (±1,0) and (0,±1), includ-
ing its interior.
(b){(x, y):|x|≤1,|y|≤1}is the square with vertices (1,1),(−1,1),(−1,−1)
and (1,−1), including its interior.

94 Bartle and Sherbert
9. For a giveny∈V
ε(x), letδ:=ε−d(x, y); thenδ>0. Show thatV δ(y)⊆V ε(x).
Sinceyis arbitrary, it follows thatV
ε(x) is an open set.
10. LetG⊆S
2be open in (S 2,d2) and letx∈f
−1
(G) so thatf(x)∈G. Then
there exists anε-neighborhoodV
ε(f(x))⊆G. Sincefis continuous atx,
there exists aδ-neighborhoodV
δ(x) such thatf(V δ(x))⊆V ε(f(x)). Since
x∈f
−1
(G) is arbitrary, we conclude thatf
−1
(G)isopenin(S 1,d1).
Iff
−1
(G)isopeninS 1for every open setG⊆S 2, and ifx 0∈S1, we let
y
0:=f(x 0)∈S 2.IfV ε(y0) is any (open)ε-neighborhood ofy 0, the hypothesis
is thatf
−1
(Vε(y0)) is an open set inS 1. Since it containsx 0, there is a
δ-neighborhoodV
δ(x0)⊆f
−1
(Vε(y0)), whencef(V δ(x0))⊆V ε(y0). Therefore
fis continuous at the arbitrary pointx
0∈S1.
11. LetG={G
α}beacoveroff(S)⊆Rby open sets inR. It follows from
11.4.11 that each setf
−1
(Gα)isopenin(S, d). Therefore, the collection
{f
−1
(Gα)}is an open cover ofS. Since (S, d) is compact, a ﬁnite subcol-
lection{f
−1
(Gα), ...,f
−1
(GαN)}coversS, whence it follows that the sets
{G
α1, ...,GαN}must form a ﬁnite subcover ofGforf(S). SinceGwas an
arbitrary open cover off(S), we conclude thatf(S) is compact.
12. Modify the proof of Theorem 11.2.4.

SELECTED GRAPHS
We include in this manual the graphs of a few selected functions that have a
special place in real analysis. They can be photocopied onto transparencies and
used with an overhead projector, if desired. They are just an indication of how
computer graphics can be used to exhibit the properties of some of the important
examples in real analysis.
The graphs were constructed using the facilities of the CALCULUS AND
MATHEMATICA project that was developed at the University of Illinois at
Urbana-Champaign by Professors H. Porta and J. J. Uhl, Jr. This is an exciting
and innovative computer laboratory in which the students use the graphic and
calculational power ofMathematicato learn calculus. The teaching of calculus
is thus brought into the modern world by having students actively interact with
modern technology.
(A)Figures 1, 2, 3. Heref(x):=x
2
sin(1/x) forx = 0 andf(0) := 0.
Three graphs are shown using three diﬀerent plot ranges, which gives a mild zoom
eﬀect. Figure 1 gives a global perspective, and shows that for “large” values ofx,
the graph approaches the liney=x. Note that
f(x)=x·
∓
sin(1/x)
1/x

and lim x→∞
∓
sin(1/x)
1/x

=1.
As the plot range becomes smaller, the oscillations become increasingly dominant.
The guiding parabolasy=x
2
andy=−x
2
are shown as dashed curves. It is
important to note that the scales on the coordinate axes are diﬀerent in the three
graphs; in Figure 1, the scale is 1 to 10, while in Figure 3 the scale is 10 to 1.
With these scales, the parabolic curves have the same shape in all three ﬁgures.
95

100
50
ﬀ50
ﬀ10 ﬀ5 510
ﬀ100
Figure 1.f(x)=x
2
sin(1/x), plot range−10<x<10
96

1
0.5
ﬀ0.5
ﬀ1 ﬀ0.5 0.5 1
ﬀ1
Figure 2.f(x)=x
2
sin(1/x), plot range−1<x<1
97

0.01
0.005
ﬀ0.005
ﬀ0.1 ﬀ0.05 0.05 0.1
ﬀ0.01
Figure 3.f(x)=x
2
sin(1/x), plot range−0.1<x<0.1
98

(B)Figures 4,5,6. Hereg(x):=x
4
[2 + sin(1/x)] forx = 0 andg(0) := 0.
There are three graphs of the functiongfor three diﬀerent plot ranges, with the
eﬀect of enlarging the graph near the origin. As the plot range decreases, the
oscillations become more apparent.
The functionghas an absolute minumum atx= 0, but the derivativeg
ﬃ
changes sign inﬁnitely often in every neighborhood ofx= 0; thus the ﬁrst deriva-
tive test does not apply. The guiding curvesy=x
4
andy=3x
4
are shown as
dashed curves. Note that the scale on the coordinate axes is diﬀerent in each
ﬁgure.
99

ﬀ0.4 ﬀ0.2 0.2 0.4
0.025
0.05
0.075
0.1
0.125
0.15
0.175
Figure 4.g(x)=x
4
[2 + sin(1/x)], plot range−0.4<x<0.4
100

ﬀ0.1 ﬀ0.05 0.05 0.1
0.00005
0.0001
0.00015
0.0002
0.00025
0.0003
Figure 5.g(x)=x
4
[2 + sin(1/x)], plot range−0.1<x<0.1
101

ﬀ0.01 ﬀ0.005 0.005 0.01
5. 10
ﬀ9
1. 10
ﬀ8
1.5 10
ﬀ8
2. 10
ﬀ8
2.5 10
ﬀ8
3. 10
ﬀ8
Figure 6.g(x)=x
4
[2 + sin(1/x)], plot range−0.01<x<0.01
102

(C)Figure 7. Hereh(x):=1/nifx=m/n, wherem, nare relatively prime
natural numbers withm<n, andh(x):=0 if 0<x<1 andxis irrational.
The graph of this function (Thomae’s function on (0, 1)) is shown. This graph
has been plotted for values ofnfrom 1 to 70, and we have decreased the level of
shading near thex-axis to keep the graph from becoming a blotch of ink for larger
values ofn. There is a gap of white just above thex-axis which must be ﬁlled in
by the imagination. No attempt is made to represent the points on thex-axis.
This function is interesting because it is discontinuous ateveryrational
number, and so has an inﬁnite number of discontinuities. However, it is Riemann
integrable.
103

0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.2 0.4 0.6 0.8 1
ﬀ0.1
Figure 7. Thomae’s function
104

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R. Bartle, D. Sherbert - Instructors Manual - Introduction to Real Analysis-John Wiley & Sons (2010).pdf