Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)

.
.
Sec?on3.7
IndeterminateformsandlHôpital’s
Rule
V63.0121.011: Calculus I
Professor Ma?hew Leingang
New York University
March 30, 2011

Announcements
IMidterm has been
returned. Please see FAQ
on Blackboard (under
”Exams and Quizzes”)
IQuiz 3 this week in
recita?on on Sec?on 2.6,
2.8, 3.1, 3.2

Objectives
IKnow when a limit is of
indeterminate form:
Iindeterminate quo?ents:
0=0,1=1
Iindeterminate products:
0B u
Iindeterminate
diﬀerences:1 1
Iindeterminate powers:
0
0
,1
0
, and 1
1
IResolve limits in
indeterminate form

Recall
Recall the limit laws from Chapter 2.
ILimit of a sum is the sum of the limits
ILimit of a diﬀerence is the diﬀerence of the limits
ILimit of a product is the product of the limits
ILimit of a quo?ent is the quo?ent of the limits ... whoops! This
is true as long as you don’t try to divide by zero.

Moreaboutdividinglimits
IWe know dividing by zero is bad.
IMost of the ?me, if an expression’s numerator approaches a
ﬁnite nonzero number and denominator approaches zero, the
quo?ent has an inﬁnite. For example:
lim
x!0
+
1
x
= +1 lim
x!0

cosx
x
3
=u

Why1=0̸=1
Consider the func?onf(x) =
1
1
x
sinx
:
.
.
x
.
y
Then lim
x!1
f(x)is of the form 1=0, but the limit does not exist and is
not inﬁnite.
Even less predictable: when numerator and denominator both go to
zero.

Experimentswithfunnylimits
Ilim
x!0
sin
2
x
x
=0
Ilim
x!0
x
sin
2
x
does not exist
Ilim
x!0
sin
2
x
sin(x
2
)
=1
Ilim
x!0
sin3x
sinx
=3
.
All of these are of the form
0
0
, and since we can get diﬀerent
answers in diﬀerent cases, we say this form isindeterminate.

LanguageNote
It depends on what the meaning of the word “is” is
IBe careful with the language here. We
arenotsaying that the limit in each
case “is”
0
0
, and therefore nonexistent
because this expression is undeﬁned.
IThe limitis of the form
0
0
, which means
we cannot evaluate it with our limit
laws.

Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.

Outline
L’Hôpital’s Rule
Rela?ve Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Diﬀerences
Indeterminate Powers

TheLinearCase
Ques?on
Iffandgare lines andf(a) =g(a) =0, what is lim
x!a
f(x)
g(x)
?
Solu?on
The func?ons f and g can be wri?en in the form
f(x) =m1(xa) g(x) =m2(xa)
So
f(x)
g(x)
=
m1
m2

TheLinearCase,Illustrated
.
. x.
y
.
y=f(x)
.
y=g(x)
..
a
..
x
.
f(x)
.
g(x)
f(x)
g(x)
=
f(x)f(a)
g(x)g(a)
=
(f(x)f(a))=(xa)
(g(x)g(a))=(xa)
=
m1
m2

Whatthen?
IBut what if the func?ons aren’t linear?
ICan we approximate a func?on near a point with a linear
func?on?
IWhat would be the slope of that linear func?on?
The
deriva?ve!

Whatthen?
IBut what if the func?ons aren’t linear?
ICan we approximate a func?on near a point with a linear
func?on?
IWhat would be the slope of that linear func?on?The
deriva?ve!

TheoremoftheDay
Theorem (L’Hopital’s Rule)
Suppose f and g are diﬀeren?able func?ons and g
′
(x)̸=0near a
(except possibly at a). Suppose that
lim
x!a
f(x) =0 and lim
x!a
g(x) =0
orlim
x!a
f(x) =1 and lim
x!a
g(x) =1
Then
lim
x!a
f(x)
g(x)
=lim
x!a
f
′
(x)
g
′
(x)
;
if the limit on the right-hand side is ﬁnite,1, oru.

MeettheMathematician
Iwanted to be a military man, but
poor eyesight forced him into
math
Idid some math on his own
(solved the “brachistocrone
problem”)
Ipaid a s?pend to Johann
Bernoulli, who proved this
theorem and named it aLer him!
Guillaume François Antoine,
Marquis de L’Hôpital
(French, 1661–1704)

Revisitingthepreviousexamples
Example
lim
x!0
sin
2
x
x
H
=lim
x!0
2sinx
.
.
sinx!0
cosx
1
=0

Revisitingthepreviousexamples
Example
lim
x!0
sin
2
x
.
.
numerator!0
sinx
2
.
.
denominator!0
H
=lim
x!0
RR2sinxcosx
.
.
numerator!0
(cosx
2
)(RR2x
.
.
denominator!0
)
H
=lim
x!0
cos
2
xsin
2
x
.
.
numerator!1
cosx
2
2x
2
sin(x
2
)
.
.
denominator!1
=1

Revisitingthepreviousexamples
Example
lim
x!0
sin
2
x
.
.
numerator!0
sinx
2.
.
denominator!0
H
=lim
x!0
RR2sinxcosx
.
.
numerator!0
(cosx
2
)(RR2x
.
.
denominator!0
)
H
=lim
x!0
cos
2
xsin
2
x
.
.
numerator!1
cosx
2
2x
2
sin(x
2
)
.
.
denominator!1
=1

Revisitingthepreviousexamples
Example
lim
x!0
sin
2
x
.
.
numerator!0
sinx
2
.
.
denominator!0
H
=lim
x!0
RR2sinxcosx
.
.
numerator!0
(cosx
2
)(RR2x
.
.
denominator!0
)
H
=lim
x!0
cos
2
xsin
2
x
.
.
numerator!1
cosx
2
2x
2
sin(x
2
)
.
.
denominator!1
=1

Revisitingthepreviousexamples
Example
lim
x!0
sin3x
sinx
H
=lim
x!0
3cos3x
cosx
=3:

AnotherExample
Example
Find
lim
x!0
x
cosx
Solu?on
The limit of the denominator is1, not0, soL’Hôpital’s rule does not
apply. The limit is0.

BewareofRedHerrings
Example
Find
lim
x!0
x
cosx
Solu?on
The limit of the denominator is1, not0, soL’Hôpital’s rule does not
apply. The limit is0.

Outline
L’Hôpital’s Rule
Rela?ve Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Diﬀerences
Indeterminate Powers

LimitsofRationalFunctions
revisited
Example
Find lim
x!1
5x
2
+3x1
3x
2
+7x+27
if it exists.
Solu?on
Using L’Hôpital:
lim
x!1
5x
2
+3x1
3x
2
+7x+27
H
=lim
x!1
10x+3
6x+7
H
=lim
x!1
10
6
=
5
3

LimitsofRationalFunctions
revisitedII
Example
Find lim
x!1
5x
2
+3x1
7x+27
if it exists.
Solu?on
Using L’Hôpital:
lim
x!1
5x
2
+3x1
7x+27
H
=lim
x!1
10x+3
7
=1

LimitsofRationalFunctions
revisitedIII
Example
Find lim
x!1
4x+7
3x
2
+7x+27
if it exists.
Solu?on
Using L’Hôpital:
lim
x!1
4x+7
3x
2
+7x+27
H
=lim
x!1
4
6x+7
=0

LimitsofRationalFunctions
Fact
Let f(x)and g(x)be polynomials of degree p and q.
IIf p>q, thenlim
x!1
f(x)
g(x)
=1
IIf p<q, thenlim
x!1
f(x)
g(x)
=0
IIf p=q, thenlim
x!1
f(x)
g(x)
is the ra?o of the leading coeﬃcients of
f and g.

Exponentialvs. geometricgrowth
Example
Find lim
x!1
e
x
x
2
, if it exists.
Solu?on
We have
lim
x!1
e
x
x
2
H
=lim
x!1
e
x
2x
H
=lim
x!1
e
x
2
=1:

Exponentialvs. geometricgrowth
Example
What about lim
x!1
e
x
x
3
?
Answer
S?ll1. (Why?)
Solu?on
lim
x!1
e
x
x
3
H
=lim
x!1
e
x
3x
2
H
=lim
x!1
e
x
6x
H
=lim
x!1
e
x
6
=1:

Exponentialvs. fractionalpowers
Example
Find lim
x!1
e
x
p
x
, if it exists.

Exponentialvs. fractionalpowers
Example
Find lim
x!1
e
x
p
x
, if it exists.
Solu?on (without L’Hôpital)
We have for all x>1, x
1=2
<x
1
, so
e
x
x
1=2
>
e
x
x
The right hand side tends to1, so the leL-hand side must, too.

Exponentialvs. fractionalpowers
Example
Find lim
x!1
e
x
p
x
, if it exists.
Solu?on (with L’Hôpital)
lim
x!1
e
x
p
x
=lim
x!1
e
x
1
2
x
1=2
=lim
x!1
2
p
xe
x
=1

Exponentialvs. anypower
Theorem
Let r be any posi?ve number. Thenlim
x!1
e
x
x
r
=1:
Proof.

Exponentialvs. anypower
Theorem
Let r be any posi?ve number. Thenlim
x!1
e
x
x
r
=1:
Proof.
Ifris a posi?ve integer, then apply L’Hôpital’s ruler?mes to the frac-
?on. You get
lim
x!1
e
x
x
r
H
=:::
H
=lim
x!1
e
x
r!
=1:

Exponentialvs. anypower
Theorem
Let r be any posi?ve number. Thenlim
x!1
e
x
x
r
=1:
Proof.
Ifris not an integer, letmbe the smallest integer greater thanr. Then
ifx>1,x
r
<x
m
;so
e
x
x
r
>
e
x
x
m
:The right-hand side tends to1by the
previous step.

Anyexponentialvs. anypower
Theorem
Let a>1and r>0. Thenlim
x!1
a
x
x
r
=1:
Proof.
Ifris a posi?ve integer, we have
lim
x!1
a
x
x
r
H
=:::
H
=lim
x!1
(lna)
r
a
x
r!
=1:
Ifrisn’t an integer, we can compare it as before.
So even lim
x!1
(1:00000001)
x
x
100000000
=1!

Logarithmicversuspowergrowth
Theorem
Let r be any posi?ve number. Thenlim
x!1
lnx
x
r
=0:
Proof.
One applica?on of L’Hôpital’s Rule here suﬃces:
lim
x!1
lnx
x
r
H
=lim
x!1
1=x
rx
r1
=lim
x!1
1
rx
r
=0:

Outline
L’Hôpital’s Rule
Rela?ve Rates of Growth
Other Indeterminate Limits
Indeterminate Products
Indeterminate Diﬀerences
Indeterminate Powers

Indeterminateproducts
Example
Find lim
x!0
+
p
xlnx
This limit is of the form 0A(u).
Solu?on
Jury-rig the expression to make an indeterminate quo?ent. Then
apply L’Hôpital’s Rule:
lim
x!0+
p
xlnx=lim
x!0
+
lnx
1=
p
x
H
=lim
x!0
+
x
1

1
2
x
3=2
=lim
x!0
+
2
p
x=0

Indeterminateproducts
Example
Find lim
x!0
+
p
xlnx
This limit is of the form 0A(u).
Solu?on
Jury-rig the expression to make an indeterminate quo?ent. Then
apply L’Hôpital’s Rule:
lim
x!0+
p
xlnx=lim
x!0
+
lnx
1=
p
x
H
=lim
x!0
+
x
1

1
2
x
3=2
=lim
x!0
+
2
p
x
=0

Indeterminateproducts
Example
Find lim
x!0
+
p
xlnx
This limit is of the form 0A(u).
Solu?on
Jury-rig the expression to make an indeterminate quo?ent. Then
apply L’Hôpital’s Rule:
lim
x!0+
p
xlnx=lim
x!0
+
lnx
1=
p
x
H
=lim
x!0
+
x
1

1
2
x
3=2
=lim
x!0
+
2
p
x=0

Indeterminatediﬀerences
Example
lim
x!0
+
(
1
x
cot2x
)
This limit is of the form1 1.

IndeterminateDiﬀerences
Solu?on
Again, rig it to make an indeterminate quo?ent.
lim
x!0
+
(
1
x
cot2x
)
=lim
x!0
+
sin(2x)xcos(2x)
xsin(2x)
H
=lim
x!0
+
cos(2x) +2xsin(2x)
2xcos(2x) +sin(2x)
=1
The limit is+1because the numerator tends to1while the
denominator tends to zero but remains posi?ve.

Checkingyourwork
This all goes in the thought cloud
lim
x!0
tan2x
2x
=1, so for smallx, tan2x2x. So cot2x
1
2x
and
1
x
cot2x
1
x

1
2x
=
1
2x
! 1
asx!0
+
.

Indeterminatepowers
Example
Find lim
x!0
+
(12x)
1=x
Solu?on

Indeterminatepowers
Example
Find lim
x!0
+
(12x)
1=x
Solu?on
Take the logarithm:
ln
(
lim
x!0
+
(12x)
1=x
)
=lim
x!0
+
ln
(
(12x)
1=x
)
=lim
x!0
+
ln(12x)
x

Indeterminatepowers
Example
Find lim
x!0
+
(12x)
1=x
Solu?on
This limit is of the form
0
0
, so we can use L’Hôpital:
lim
x!0
+
ln(12x)
x
H
=lim
x!0
+
2
12x
1
=2

Indeterminatepowers
Example
Find lim
x!0
+
(12x)
1=x
Solu?on
This limit is of the form
0
0
, so we can use L’Hôpital:
lim
x!0
+
ln(12x)
x
H
=lim
x!0
+
2
12x
1
=2
This is not the answer, it’s the log of the answer! So the answer we
want ise
2
.

Anotherindeterminatepowerlimit
Example
Find lim
x!0
+
(3x)
4x
Solu?on
ln lim
x!0
+
(3x)
4x
=lim
x!0
+
ln(3x)
4x
=lim
x!0
+
4xln(3x) =lim
x!0
+
ln(3x)
1=4x
H
=lim
x!0
+
3=3x
1=4x
2
=lim
x!0
+
(4x) =0
So the answer is e
0
=1.

Summary
FormMethod
0
0
L’Hôpital’s rule directly
1
1
L’Hôpital’s rule directly
0A ujiggle to make
0
0
or
1
1
.
1 1combine to make an indeterminate product or quo?ent
0
0
take ln to make an indeterminate product
1
0
di?o
1
1
di?o

FinalThoughts
IL’Hôpital’s Rule only works on indeterminate quo?ents
ILuckily, most indeterminate limits can be transformed into
indeterminate quo?ents
IL’Hôpital’s Rule gives wrong answers for non-indeterminate
limits!

Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)

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Lesson 17: Indeterminate forms and l'Hôpital's Rule (slides)