Forms study purpusesChapter-2-Limits.pdf
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About This Presentation
To find the limits in calculator of calculus it can gives deal to understand easy the calculator on mathematics it's help the students to practice the problem solving in this module do that you can understand easy for school tomorrow, but you can understand because student loans in forbearance a...
Slide Content
Overview:
This chapter introduces concepts on functions and limits. The purpose
of this module is to impart a thorough knowledge and understanding of these
concepts.
Learning Outcomes:
After working with the module, you will be able to:
1. Illustrate the limit of a function using a table of values and the graph of a
function.
2. Distinguish between lim
?????? →�
�(??????)and f(a)
3. Understand limits and its theorems.
4. Apply the limit laws in evaluating the limit of algebraic functions.
5. Illustrate the continuity of a function
6. Determine whether the function is continuous at a number or not.
Familiarity with the limit concept is absolutely essential for a deeper
understanding of calculus. In this section, we shall begin our discussion of the
limit of a function and its theorems.
Work on the following activity for recall and practice.
To understand the concepts of limits, let us consider the function defined
by the equation
f(x) = 3x + 1
and assign some values to x near, but not equal to a specific value, say 3. We
construct a table of values shown here:
CHAPTER 2
LIMITS
2.1 LIMITS OF FUNCTIONS
Let f(x) = 3x -1. Fill in the table below by calculating the values of f using the
pre assigned values of x.
x -2 -1 0 1 3
f(x)= 3x – 1
This chapter introduces concepts on functions and limits. The purpose
of this module is to impart a thorough knowledge and understanding of these
concepts.
Learning Outcomes:
After working with the module, you will be able to:
1. Illustrate the limit of a function using a table of values and the graph of a
function.
2. Distinguish between lim
?????? →�
�(??????)and f(a)
3. Understand limits and its theorems.
4. Apply the limit laws in evaluating the limit of algebraic functions.
5. Illustrate the continuity of a function
6. Determine whether the function is continuous at a number or not.
Familiarity with the limit concept is absolutely essential for a deeper
understanding of calculus. In this section, we shall begin our discussion of the
limit of a function and its theorems.
Work on the following activity for recall and practice.
To understand the concepts of limits, let us consider the function defined
by the equation
f(x) = 3x + 1
and assign some values to x near, but not equal to a specific value, say 3. We
construct a table of values shown here:
CHAPTER 2
LIMITS
2.1 LIMITS OF FUNCTIONS
Let f(x) = 3x -1. Fill in the table below by calculating the values of f using the
pre assigned values of x.
x -2 -1 0 1 3
f(x)= 3x – 1
13
x 2.500 2.670 2.900 3.005 3.230 3.300
f(x) 8.500 9.01 9.7 10.015 10.69 10.9
The table shows that when x is near 3, whether a little less or a little
greater than 3, f(x) = 3x + 1 = 10. In other words, “3x + 1 approaches the
number 10 as a limit when x approaches 3”. The abbreviated symbolic form for
this statement is
3x + 1 → 10 as x → 3
We may also say that “the limit of 3x + 1 as x approaches 3 is 10.” In symbol, we
write this as
lim
??????→3
3??????+1 = 10
Let us consider lim (1 - 3x)
x→ 2
To evaluate the given limit, we will make use of a table to help us keep
track of the effect that the approach of x towards 2 will have on f(x). We first
consider approaching 2 from its left or values less than 2.
x 1.0 1.4 1.6 1.8 1.99 1.99999
f(x) -2 -3.2 -3.8 -4.4 -4.97 -4.99997
x 3.0 2.8 2.6 2.1 2.005 2.00001
f(x) 8 7.4 6.8 5.3 5.015 5.00003
Observe that as the values of x get closer and closer to 2, the values of
f(x) get closer and closer to 5. This behavior can be shown that no matter what
set of values, or what direction is taken in approaching 2. In symbols,
lim
?????? → 2
(1−3??????) = 5
ILLUSTRATION 1:
Definition. Let f(x) be any function and let a and L be numbers. If we can make f(x) as
close to L as we please by choosing x sufficiently close to a then we say that the limit
of f(x) as x approaches a is L or symbolically.
lim
??????→ �
�(??????) = L
x 2.500 2.670 2.900 3.005 3.230 3.300
f(x) 8.500 9.01 9.7 10.015 10.69 10.9
The table shows that when x is near 3, whether a little less or a little
greater than 3, f(x) = 3x + 1 = 10. In other words, “3x + 1 approaches the
number 10 as a limit when x approaches 3”. The abbreviated symbolic form for
this statement is
3x + 1 → 10 as x → 3
We may also say that “the limit of 3x + 1 as x approaches 3 is 10.” In symbol, we
write this as
lim
??????→3
3??????+1 = 10
Let us consider lim (1 - 3x)
x→ 2
To evaluate the given limit, we will make use of a table to help us keep
track of the effect that the approach of x towards 2 will have on f(x). We first
consider approaching 2 from its left or values less than 2.
x 1.0 1.4 1.6 1.8 1.99 1.99999
f(x) -2 -3.2 -3.8 -4.4 -4.97 -4.99997
x 3.0 2.8 2.6 2.1 2.005 2.00001
f(x) 8 7.4 6.8 5.3 5.015 5.00003
Observe that as the values of x get closer and closer to 2, the values of
f(x) get closer and closer to 5. This behavior can be shown that no matter what
set of values, or what direction is taken in approaching 2. In symbols,
lim
?????? → 2
(1−3??????) = 5
ILLUSTRATION 1:
Definition. Let f(x) be any function and let a and L be numbers. If we can make f(x) as
close to L as we please by choosing x sufficiently close to a then we say that the limit
of f(x) as x approaches a is L or symbolically.
lim
??????→ �
�(??????) = L
14
Investigate lim /x/ through a table of values
x→0
X /x/
-0.3
-0.01
-0.00009
-0.00000001
0.3
0.01
0.00009
0.00000001
Hence, lim /x/ = 0
x→0
To further analyze the idea presented about limits, let us illustrate the
graph of f(x) =
(??????+3)(??????−2)
??????−2
as shown in the figure.
Figure 1
The graph of f is a straight line defined at every real number x except at
2. It is undefined at x = 2, but let us observe the values that f takes as x becomes
closer and closer to 2, and not the value of 2.
The graph of f can be seen in figure1 that as x becomes closer to 2, the
corresponding values of f become closer and closer to 5.
ILLUSTRATION 2:
ILLUSTRATION 3:
The graph of this
function was sketch
through Desmos.
https://www.desmos.
com/calculator
Investigate lim /x/ through a table of values
x→0
X /x/
-0.3
-0.01
-0.00009
-0.00000001
0.3
0.01
0.00009
0.00000001
Hence, lim /x/ = 0
x→0
To further analyze the idea presented about limits, let us illustrate the
graph of f(x) =
(??????+3)(??????−2)
??????−2
as shown in the figure.
Figure 1
The graph of f is a straight line defined at every real number x except at
2. It is undefined at x = 2, but let us observe the values that f takes as x becomes
closer and closer to 2, and not the value of 2.
The graph of f can be seen in figure1 that as x becomes closer to 2, the
corresponding values of f become closer and closer to 5.
ILLUSTRATION 2:
ILLUSTRATION 3:
The graph of this
function was sketch
through Desmos.
https://www.desmos.
com/calculator
15
Investigate through a table of values
lim f(x) f(x) = x + 1 if x < 4
x→4 (x-4)
2
+ 3 if x≥ 4
This looks a bit different, but the logic and procedure are exactly the
same. We still approach the constant 4 from the left and from the light, but note
that we should evaluate the appropriate corresponding functional expression.
In this case, when x approaches 4 from the left, the values taken should be
substituted in f(x) = x + 1. This is the part of the function which accepts values
less than 4.
x f(x)
3.7
3.85
3.995
3.99999
4.7
4.85
4.995
4.99999
On the other hand, when x approaches 4 from the right, the values taken
should be substituted in f(x) = (x - 4)
2
+ 3.
x f(x)
4.3
4.1
4.001
4.00001
3.09
3.01
3.000001
3.0000000001
Observe that the values that f(x) approaches are not equal, namely, f(x)
approaches 5 from the left while it approaches 3 from the right. In such case, we
say that the limit of the given function does not exist.
Remark:
In general, if the values of f(x) approach different values as x approaches
a from the left and from the right, the limit does not exist. This is one the criteria
to determine if a limit does not exist.
We define the limit of f(x) as x approaches a from the right as the number
L to which f(x) can be made as close as we wish by making x close enough to a,
from the right of a (that is, x > a), if such number exists. In case, we write
lim
?????? → �
+
�(??????)=??????
ILLUSTRATION 4:
Investigate through a table of values
lim f(x) f(x) = x + 1 if x < 4
x→4 (x-4)
2
+ 3 if x≥ 4
This looks a bit different, but the logic and procedure are exactly the
same. We still approach the constant 4 from the left and from the light, but note
that we should evaluate the appropriate corresponding functional expression.
In this case, when x approaches 4 from the left, the values taken should be
substituted in f(x) = x + 1. This is the part of the function which accepts values
less than 4.
x f(x)
3.7
3.85
3.995
3.99999
4.7
4.85
4.995
4.99999
On the other hand, when x approaches 4 from the right, the values taken
should be substituted in f(x) = (x - 4)
2
+ 3.
x f(x)
4.3
4.1
4.001
4.00001
3.09
3.01
3.000001
3.0000000001
Observe that the values that f(x) approaches are not equal, namely, f(x)
approaches 5 from the left while it approaches 3 from the right. In such case, we
say that the limit of the given function does not exist.
Remark:
In general, if the values of f(x) approach different values as x approaches
a from the left and from the right, the limit does not exist. This is one the criteria
to determine if a limit does not exist.
We define the limit of f(x) as x approaches a from the right as the number
L to which f(x) can be made as close as we wish by making x close enough to a,
from the right of a (that is, x > a), if such number exists. In case, we write
lim
?????? → �
+
�(??????)=??????
ILLUSTRATION 4:
16
Similarly, we define the limit of f(x) as x approaches a from the left by
lim
?????? → �
+
�(??????)=??????
We can think that the limit is a two-sided limit. Indeed, the limit of f(x)
as x approaches a exists if and only if the one-sided limit of f(x) as x approaches
from the right and from the left both exist and are equal. We will discuss more
about one-sided limits in lesson 1.5
Use the tabular and graphical approaches to evaluate lim
??????→5
(??????
2
−6?????? +11).
Tabular approach:
The following tables show the values of f for some values of x that are
close to 3.
x f(x)= ??????
2
−6?????? +
10
4 3
4.5 4.25
4.9 5.61
4.99 5.9601
4.999 5.996001
4.999999 5.999996
From the table above, we conclude that the lim
??????→5
(??????
2
−6?????? +11) = 6
Graphical Approach:
Figure 2
Example 1
x f(x)= ??????
2
−6?????? +
1
6 11
5.5 8.25
5.01 6.04
5.001 6.004
5.0001 6.0004
5.000001 6.000004
f(x) 6
x→ 5
Similarly, we define the limit of f(x) as x approaches a from the left by
lim
?????? → �
+
�(??????)=??????
We can think that the limit is a two-sided limit. Indeed, the limit of f(x)
as x approaches a exists if and only if the one-sided limit of f(x) as x approaches
from the right and from the left both exist and are equal. We will discuss more
about one-sided limits in lesson 1.5
Use the tabular and graphical approaches to evaluate lim
??????→5
(??????
2
−6?????? +11).
Tabular approach:
The following tables show the values of f for some values of x that are
close to 3.
x f(x)= ??????
2
−6?????? +
10
4 3
4.5 4.25
4.9 5.61
4.99 5.9601
4.999 5.996001
4.999999 5.999996
From the table above, we conclude that the lim
??????→5
(??????
2
−6?????? +11) = 6
Graphical Approach:
Figure 2
Example 1
x f(x)= ??????
2
−6?????? +
1
6 11
5.5 8.25
5.01 6.04
5.001 6.004
5.0001 6.0004
5.000001 6.000004
f(x) 6
x→ 5
17
We sketch the graph using Desmos which is a parabola opening upwards
and with vertex (3,2). We can see in the above figure that using the values of x
becoming closer to 5, the corresponding y-coordinates or f(x) values are
becoming closer and closer to 6.
This section deals with several theorems by means of which we shall be
able to evaluate the limits of functions rapidly and efficiently. To evaluate or to
find
lim f(x)
x → a
means that we are to find the number L that f(x) is near, whenever x is
near a but not equal to a. Of course, when x = a, the value of the function is f(a).
It may be that f(a) is also the limit, that is, L=f(a).
Thus, to evaluate
lim (4 - x
2
)
x→1
means to find a number which 4 -x
2
is near whenever x is near the
number 1. By definition 1.1, we know that
lim (4 - x
2
) = 3
x→ 1
Self-Assessment Questions 1.1
2.2 THEOREMS ON LIMITS
1. Complete the following tables of values to investigate lim
?????? →0
??????−1
??????+1
x f(x) x f(x)
-1
-0.8
-0.35
-0.1
-0.09
-0.003
-0.000001
1
0.75
0.45
0.2
0.09
0.0003
0.000001
2. Construct a table of values to investigate:
lim
?????? →2
2??????+1
??????−3
We sketch the graph using Desmos which is a parabola opening upwards
and with vertex (3,2). We can see in the above figure that using the values of x
becoming closer to 5, the corresponding y-coordinates or f(x) values are
becoming closer and closer to 6.
This section deals with several theorems by means of which we shall be
able to evaluate the limits of functions rapidly and efficiently. To evaluate or to
find
lim f(x)
x → a
means that we are to find the number L that f(x) is near, whenever x is
near a but not equal to a. Of course, when x = a, the value of the function is f(a).
It may be that f(a) is also the limit, that is, L=f(a).
Thus, to evaluate
lim (4 - x
2
)
x→1
means to find a number which 4 -x
2
is near whenever x is near the
number 1. By definition 1.1, we know that
lim (4 - x
2
) = 3
x→ 1
Self-Assessment Questions 1.1
2.2 THEOREMS ON LIMITS
1. Complete the following tables of values to investigate lim
?????? →0
??????−1
??????+1
x f(x) x f(x)
-1
-0.8
-0.35
-0.1
-0.09
-0.003
-0.000001
1
0.75
0.45
0.2
0.09
0.0003
0.000001
2. Construct a table of values to investigate:
lim
?????? →2
2??????+1
??????−3
18
since by choosing x sufficiently close to 1, (4 – x
2)
come as close to 3 as
we please.
We shall use the following theorems which we shall state symbolically
without proof.
a. lim
?????? →2
12 = 12
b. lim
??????→5
7?????? = 35
a. If lim
?????? → �
�(??????) = 7 and lim
?????? → �
�(??????) = 5, then
lim
?????? → �
3�(??????) =(3)( 7)= 21 and lim
?????? → �
8�(??????) = (8)(5)= 40,
Example 2
Example 3
Theorem 1. Basic Limit Laws
lim c = c c= any constant a= any real number
x → a
lim
?????? → �
�?????? = ca
Theorem 2. Limit of a Constant Multiple
If c is any constant and lim
?????? → �
�(??????)=??????, then
lim c f(x) = c lim f(x)= cL
x →a x →a
Theorem 3. Limit of a Sum and Difference
lim [f(x) + g(x)] = lim f(x) + lim g(x)
x →a x →a x →a
lim [f(x) - g(x)] = lim f(x) - lim g(x)
x →a x →a x →a
since by choosing x sufficiently close to 1, (4 – x
2)
come as close to 3 as
we please.
We shall use the following theorems which we shall state symbolically
without proof.
a. lim
?????? →2
12 = 12
b. lim
??????→5
7?????? = 35
a. If lim
?????? → �
�(??????) = 7 and lim
?????? → �
�(??????) = 5, then
lim
?????? → �
3�(??????) =(3)( 7)= 21 and lim
?????? → �
8�(??????) = (8)(5)= 40,
Example 2
Example 3
Theorem 1. Basic Limit Laws
lim c = c c= any constant a= any real number
x → a
lim
?????? → �
�?????? = ca
Theorem 2. Limit of a Constant Multiple
If c is any constant and lim
?????? → �
�(??????)=??????, then
lim c f(x) = c lim f(x)= cL
x →a x →a
Theorem 3. Limit of a Sum and Difference
lim [f(x) + g(x)] = lim f(x) + lim g(x)
x →a x →a x →a
lim [f(x) - g(x)] = lim f(x) - lim g(x)
x →a x →a x →a
19
a. If lim
?????? → �
�(??????) = 9 and lim
?????? → �
�(??????) = 6, then
lim
?????? → �
�(??????) + lim
?????? → �
�(??????) = 9 + 6 = 15
lim
?????? → �
�(??????) - lim
?????? → �
�(??????) = 9 - 6 = 3
b. If lim
?????? → �
�(??????) =−3 and lim
?????? → �
�(??????) =
1
6
, then
lim
?????? → �
3�(??????) + lim
?????? → �
6�(??????) = 3(-3) + 6(
1
6
) = -1 + 1 = 0
lim
?????? → �
5�(??????) - lim
?????? → �
3�(??????) = 5(-3) - 3(
1
6
) = -15 -
1
2
=
−31
2
a. If lim
?????? → �
�(??????) = 8 and lim
?????? → �
�(??????) = 7, then
lim
?????? → �
�(??????)�(??????) = lim
?????? → �
�(??????)∙lim
?????? → �
�(??????)= (8)(7) = 56
b. If lim
?????? → �
�(??????) =−5 and lim
?????? → �
�(??????) =
1
3
, then what is the value of
lim
?????? → �
[2�(??????)][3 �(??????)]?
lim
?????? → �
[2�(??????)][3 �(??????)]=lim
?????? → �
2�(??????)∙ lim
?????? → �
3�(??????) = 2(-5) [ 3(
1
3
) ]= -10
a. If lim
?????? → �
�(??????) = 10 and lim
?????? → �
�(??????) = 5, then
Example 4
Example 5
Example 6
Theorem 4. Limit of a Product
lim [ f(x)g(x)] = lim f(x) . lim g(x)
x →a x →a x →a
Theorem 5. Limit of a Quotient
lim
?????? → �
�(??????)
�(??????)
=
lim
?????? → ??????
�(??????)
lim
?????? → ??????
�(??????)
Where g(x) ≠ 0
a. If lim
?????? → �
�(??????) = 9 and lim
?????? → �
�(??????) = 6, then
lim
?????? → �
�(??????) + lim
?????? → �
�(??????) = 9 + 6 = 15
lim
?????? → �
�(??????) - lim
?????? → �
�(??????) = 9 - 6 = 3
b. If lim
?????? → �
�(??????) =−3 and lim
?????? → �
�(??????) =
1
6
, then
lim
?????? → �
3�(??????) + lim
?????? → �
6�(??????) = 3(-3) + 6(
1
6
) = -1 + 1 = 0
lim
?????? → �
5�(??????) - lim
?????? → �
3�(??????) = 5(-3) - 3(
1
6
) = -15 -
1
2
=
−31
2
a. If lim
?????? → �
�(??????) = 8 and lim
?????? → �
�(??????) = 7, then
lim
?????? → �
�(??????)�(??????) = lim
?????? → �
�(??????)∙lim
?????? → �
�(??????)= (8)(7) = 56
b. If lim
?????? → �
�(??????) =−5 and lim
?????? → �
�(??????) =
1
3
, then what is the value of
lim
?????? → �
[2�(??????)][3 �(??????)]?
lim
?????? → �
[2�(??????)][3 �(??????)]=lim
?????? → �
2�(??????)∙ lim
?????? → �
3�(??????) = 2(-5) [ 3(
1
3
) ]= -10
a. If lim
?????? → �
�(??????) = 10 and lim
?????? → �
�(??????) = 5, then
Example 4
Example 5
Example 6
Theorem 4. Limit of a Product
lim [ f(x)g(x)] = lim f(x) . lim g(x)
x →a x →a x →a
Theorem 5. Limit of a Quotient
lim
?????? → �
�(??????)
�(??????)
=
lim
?????? → ??????
�(??????)
lim
?????? → ??????
�(??????)
Where g(x) ≠ 0
20
lim
?????? → �
�(??????)
�(??????)
=
lim
?????? → ??????
�(??????)
lim
?????? → ??????
�(??????)
=
10
5
= 2
b. If lim
?????? → �
�(??????) =−5 and lim
?????? → �
�(??????) = 10, then
lim
?????? → �
8�(??????)
�(??????)
=
lim
?????? → ??????
8�(??????)
lim
?????? → ??????
�(??????)
=
8(−5)
10
=
−40
10
=- 4
a. If lim
?????? →�
�(??????) = 7 , then
lim
?????? →�
�(??????)
4
= 7
4
= 2401
b. If lim
?????? →�
�(??????) =
−1
3
, then
lim
?????? →�
�(??????)
2
= (
−1
3
)
2
=
1
9
a. If lim
?????? →�
�(??????) = 64 , then
lim
?????? →�
√�(??????) = √lim
?????? →�
�(??????) = √64 = 8
Example 7
Example 7
Theorem 6. Limit of a Power
If n is any positive integer, then
lim [f(x)]
n = [lim f(x)]
n
x→a x→a
Theorem 7. Limit of a Radical Function
lim √�(??????)
??????
= √�??????� �(??????)
??????
n = any positive integer and f(x) ≥0 if n is even
x →a x →a
lim
?????? → �
�(??????)
�(??????)
=
lim
?????? → ??????
�(??????)
lim
?????? → ??????
�(??????)
=
10
5
= 2
b. If lim
?????? → �
�(??????) =−5 and lim
?????? → �
�(??????) = 10, then
lim
?????? → �
8�(??????)
�(??????)
=
lim
?????? → ??????
8�(??????)
lim
?????? → ??????
�(??????)
=
8(−5)
10
=
−40
10
=- 4
a. If lim
?????? →�
�(??????) = 7 , then
lim
?????? →�
�(??????)
4
= 7
4
= 2401
b. If lim
?????? →�
�(??????) =
−1
3
, then
lim
?????? →�
�(??????)
2
= (
−1
3
)
2
=
1
9
a. If lim
?????? →�
�(??????) = 64 , then
lim
?????? →�
√�(??????) = √lim
?????? →�
�(??????) = √64 = 8
Example 7
Example 7
Theorem 6. Limit of a Power
If n is any positive integer, then
lim [f(x)]
n = [lim f(x)]
n
x→a x→a
Theorem 7. Limit of a Radical Function
lim √�(??????)
??????
= √�??????� �(??????)
??????
n = any positive integer and f(x) ≥0 if n is even
x →a x →a
21
b. a. If lim
?????? →�
�(??????) = 8 , then
lim
?????? →�
√�(??????)
3
= √lim
?????? →�
�(??????)
3
=√8
3
= 2
This lesson discusses how to evaluate the limits of algebraic functions,
particularly polynomial, rational and radical functions.
In the case of polynomials, evaluating limits involves direct substitution
for every variable x in the given polynomial function. The basic laws in limits
should be applied in evaluating polynomial functions.
Self-Assessment Questions 1.2
2. 3 LIMITS OF ALGEBRAIC FUNCTIONS
A. Apply the limit laws to evaluate the following limits.
1. lim
?????? → 2
35 2. lim
??????→ 4
2?????? 3. lim
??????→ −1
3??????−1
B. Given lim
??????→ �
�(??????)= 16 and lim
??????→ �
�(??????) = 4, use the limit to evaluate the
following:
1. lim
??????→ �
4�(??????) 4. lim
??????→�
2�(??????) + lim
??????→�
�(??????)
2. lim
??????→ �
5�(??????) 5. lim
??????→�
�(??????)�(??????)
3. lim
??????→ 4
√�(??????)
Definition of a Polynomial Function
A polynomial function f in terms of x in the form
f(x) = anx
n + an-1x
n-1 + …+a2x
2 + a1x + a0
where ai, for i= 0, 1, 2…, n, are real coefficients and n is a nonnegative integer.
If the leading coefficients an ≠ 0, then n is the degree of the polynomial.
b. a. If lim
?????? →�
�(??????) = 8 , then
lim
?????? →�
√�(??????)
3
= √lim
?????? →�
�(??????)
3
=√8
3
= 2
This lesson discusses how to evaluate the limits of algebraic functions,
particularly polynomial, rational and radical functions.
In the case of polynomials, evaluating limits involves direct substitution
for every variable x in the given polynomial function. The basic laws in limits
should be applied in evaluating polynomial functions.
Self-Assessment Questions 1.2
2. 3 LIMITS OF ALGEBRAIC FUNCTIONS
A. Apply the limit laws to evaluate the following limits.
1. lim
?????? → 2
35 2. lim
??????→ 4
2?????? 3. lim
??????→ −1
3??????−1
B. Given lim
??????→ �
�(??????)= 16 and lim
??????→ �
�(??????) = 4, use the limit to evaluate the
following:
1. lim
??????→ �
4�(??????) 4. lim
??????→�
2�(??????) + lim
??????→�
�(??????)
2. lim
??????→ �
5�(??????) 5. lim
??????→�
�(??????)�(??????)
3. lim
??????→ 4
√�(??????)
Definition of a Polynomial Function
A polynomial function f in terms of x in the form
f(x) = anx
n + an-1x
n-1 + …+a2x
2 + a1x + a0
where ai, for i= 0, 1, 2…, n, are real coefficients and n is a nonnegative integer.
If the leading coefficients an ≠ 0, then n is the degree of the polynomial.
22
Evaluate the following limits:
1. lim (x
2
+3x + 4) = (2)2 + 3(2) +4 = 14
x→2
2. lim (3x + 5)
2
= 3(2) + 5
2
= 169
x→2
3. lim (4x + 5) √?????? + 6 = [(4)3 + 5] (√9 ) = 17(3) = 51
x→3
The laws of limits are useful in evaluating limits of rational functions. It
also involves direct substitution of a on very variable x provided that it will not
make the function undefined.
1. lim
??????→ 2
??????
?????? + 2
=
2
2+ 2
=
1
2
2. lim
??????→0
??????
2
−24
?????? + 6
=
0
2
−24
0 + 6
=
−24
6
= -4
3. lim
?????? →1
??????
2
+2?????? + 1
2??????
2
+ 4
=
1
2
+2(1) + 1
2(1)
2
+ 4
=
4
6
=
2
3
Example 8
Example 9
Definition of a Rational Function
A rational function f in terms of x is of the form
f(x) =
�(??????)
�(??????)
where p(x) and q(x) are polynomials in x and q(x)≠ 0
Theorem 8. Limit of a Rational Function
For a rational function f in terms of x,
lim
??????→ �
�(??????) = lim
??????→�
�(??????)
�(??????)
=
�(�)
�(�)
Where a is any number and q(a)≠ 0
Evaluate the following limits:
1. lim (x
2
+3x + 4) = (2)2 + 3(2) +4 = 14
x→2
2. lim (3x + 5)
2
= 3(2) + 5
2
= 169
x→2
3. lim (4x + 5) √?????? + 6 = [(4)3 + 5] (√9 ) = 17(3) = 51
x→3
The laws of limits are useful in evaluating limits of rational functions. It
also involves direct substitution of a on very variable x provided that it will not
make the function undefined.
1. lim
??????→ 2
??????
?????? + 2
=
2
2+ 2
=
1
2
2. lim
??????→0
??????
2
−24
?????? + 6
=
0
2
−24
0 + 6
=
−24
6
= -4
3. lim
?????? →1
??????
2
+2?????? + 1
2??????
2
+ 4
=
1
2
+2(1) + 1
2(1)
2
+ 4
=
4
6
=
2
3
Example 8
Example 9
Definition of a Rational Function
A rational function f in terms of x is of the form
f(x) =
�(??????)
�(??????)
where p(x) and q(x) are polynomials in x and q(x)≠ 0
Theorem 8. Limit of a Rational Function
For a rational function f in terms of x,
lim
??????→ �
�(??????) = lim
??????→�
�(??????)
�(??????)
=
�(�)
�(�)
Where a is any number and q(a)≠ 0
23
For some rational functions, evaluating their limits as x approaches the
number a may not be a direct substitution if after evaluation f(a) leads to
0
0
which is not defined and is called an indeterminate form.
Suppose at x = a, p(a) = q(a) = 0
f(a) = p(a) = 0
q(a) 0
Which is undefined. We say at x =a, the function f(x) assumes in the
indeterminate form
0
0
. We shall in the examples below that even if f(x) assumes
the indeterminate forms 0/0 at x=a, the limit of f(x) may be definite, that is, the
limit exists.
The most common forms of indeterminates are:
0
0
,
∞
∞
1. Evaluate lim x
2
- 4
x→2 x – 2
Self-Assessment Questions 1.3
2.4 INDETERMINATE FORMS
Example 9
Consider the function defined by
f(x)= p(x)
q(x)
Where q(x) ≠ 0
Evaluate the following limits:
1. lim
?????? →−2
4??????
2
−5?????? + 1
2. lim
?????? →−1
6??????
2
− 1
3. lim
?????? →4
??????
2
−??????−6
??????−3
4. lim
?????? →2
√?????? + 3 −2
??????−1
For some rational functions, evaluating their limits as x approaches the
number a may not be a direct substitution if after evaluation f(a) leads to
0
0
which is not defined and is called an indeterminate form.
Suppose at x = a, p(a) = q(a) = 0
f(a) = p(a) = 0
q(a) 0
Which is undefined. We say at x =a, the function f(x) assumes in the
indeterminate form
0
0
. We shall in the examples below that even if f(x) assumes
the indeterminate forms 0/0 at x=a, the limit of f(x) may be definite, that is, the
limit exists.
The most common forms of indeterminates are:
0
0
,
∞
∞
1. Evaluate lim x
2
- 4
x→2 x – 2
Self-Assessment Questions 1.3
2.4 INDETERMINATE FORMS
Example 9
Consider the function defined by
f(x)= p(x)
q(x)
Where q(x) ≠ 0
Evaluate the following limits:
1. lim
?????? →−2
4??????
2
−5?????? + 1
2. lim
?????? →−1
6??????
2
− 1
3. lim
?????? →4
??????
2
−??????−6
??????−3
4. lim
?????? →2
√?????? + 3 −2
??????−1
24
Solution:
By straight substitution, we will have
x
2
– 4 = 4 – 4 = 0
x – 2 2 – 2 0
which is meaningless. That is, at x = 2, the function assumes the
indeterminate form
0
0
.
However, if x≠ 2, then by factoring of the numerator
x
2
– 4 = (x-2)(x+2) = x+ 2
x – 2 = (x-2)
Therefore, to evaluate the limits of the given function, we proceed as
follows:
lim
??????
2
− 4
?????? − 2
= lim (x – 2) ( x + 2)
x→ 2 x →2 x -2
= lim (x+ 2)
x→2
= 2 + 2 = 4
2. Evaluate the limit: lim x
2
+ 2x – 8
x→2 3x – 6
Solution:
lim x
2
+ 2x – 8 = 4 + 4 – 8 = 0 Indeterminate form!
x→2 3x – 6 3(2) – 6 0
Hence,
lim (x + 4)(x-2) = (x+4) = (2 + 4) = 6 = 2
x → 2 3(x-2) 3 3 3
Solution:
By straight substitution, we will have
x
2
– 4 = 4 – 4 = 0
x – 2 2 – 2 0
which is meaningless. That is, at x = 2, the function assumes the
indeterminate form
0
0
.
However, if x≠ 2, then by factoring of the numerator
x
2
– 4 = (x-2)(x+2) = x+ 2
x – 2 = (x-2)
Therefore, to evaluate the limits of the given function, we proceed as
follows:
lim
??????
2
− 4
?????? − 2
= lim (x – 2) ( x + 2)
x→ 2 x →2 x -2
= lim (x+ 2)
x→2
= 2 + 2 = 4
2. Evaluate the limit: lim x
2
+ 2x – 8
x→2 3x – 6
Solution:
lim x
2
+ 2x – 8 = 4 + 4 – 8 = 0 Indeterminate form!
x→2 3x – 6 3(2) – 6 0
Hence,
lim (x + 4)(x-2) = (x+4) = (2 + 4) = 6 = 2
x → 2 3(x-2) 3 3 3
25
In our previous discussion, we evaluated limits of a functions as x
approaches a particular number a by taking the values of x from the left of a and
values from right of a. The limits exists if the resulting function value from both
sides approach the same number L.
Consider to evaluate lim
??????→ 0
√??????+ 1 which is undefined when x < 0, thus, it
limits as it approaches to 0 through numbers less than 0 will not exist. However,
if we only use the values of x > 0, the f(x) may be made close to 1.
This leads us to consider separate evaluation of limit for cases when x
approaches a through numbers greater than a and through numbers less than
a. This is now considered as evaluation of one-sided limits which may be left-
hand side limits or a right-hand side limit.
When we make x approaches to a coming from the left, we use values
that are less than a but close to the value of a. we call this, the left-hand side of f
as x approaches to a.
Self-Assessment Questions 1.4
2.5 ONE- SIDED LIMITS
Evaluate the following limits:
1. lim x
3 - 64
x→4 x
2 – 16
2. lim x
3 -13x + 12
x→3 x
3 – 14x + 15
3. lim (x+3)
2 - 9
x→0 2x
Definition of Left-hand side Limit.
Let f be a function defined at every number in some open interval (b,a). Then
the limit of f(x) as x approaches a from the left side L written as
lim
?????? →�
−
�(??????) = ??????
In our previous discussion, we evaluated limits of a functions as x
approaches a particular number a by taking the values of x from the left of a and
values from right of a. The limits exists if the resulting function value from both
sides approach the same number L.
Consider to evaluate lim
??????→ 0
√??????+ 1 which is undefined when x < 0, thus, it
limits as it approaches to 0 through numbers less than 0 will not exist. However,
if we only use the values of x > 0, the f(x) may be made close to 1.
This leads us to consider separate evaluation of limit for cases when x
approaches a through numbers greater than a and through numbers less than
a. This is now considered as evaluation of one-sided limits which may be left-
hand side limits or a right-hand side limit.
When we make x approaches to a coming from the left, we use values
that are less than a but close to the value of a. we call this, the left-hand side of f
as x approaches to a.
Self-Assessment Questions 1.4
2.5 ONE- SIDED LIMITS
Evaluate the following limits:
1. lim x
3 - 64
x→4 x
2 – 16
2. lim x
3 -13x + 12
x→3 x
3 – 14x + 15
3. lim (x+3)
2 - 9
x→0 2x
Definition of Left-hand side Limit.
Let f be a function defined at every number in some open interval (b,a). Then
the limit of f(x) as x approaches a from the left side L written as
lim
?????? →�
−
�(??????) = ??????
26
When we make x approaches to a coming from the right, we use values
that are greater than a but close to the value of a. we call this, the right-hand
side of f as x approaches to a.
1. Evaluate lim
??????→5
−
√5− ?????? and lim
??????→5
+
√5−??????
Solution:
When the values are coming from the right side of 5, f(x) = √5− ?????? is not
defined. Therefore, lim
??????→5
+
√5−?????? does not exist.
When the values are coming from the left side of 5, f(x) = √5− ?????? is
defined. Therefore, lim
??????→5
−
√5−?????? = √5−4 = 1
1. Consider the function f defined by f(x) =
3,??????>1
1,??????=1
−2??????<1
. Evaluate each of the
following limits, if it exists.
a. lim
??????→1
+
�(??????) b. lim
??????→1
−
�(??????) c. lim
??????→1
�(??????)
Example 10
Example 11
Definition of Right-hand side Limit.
Let f be a function defined at every number in some open interval (a,d). Then
the limit of f(x) as x approaches a from the left side L written as
lim
?????? →�
+
�(??????) = ??????
Theorem 9. Existence of Limit
The limit of f as x approaches a exists and is equal to the number L if and only if
lim
?????? →�
−
�(??????) ��� lim
??????→�
+
�(??????) both exist and are equal to L. That is,
lim
??????→�
�(??????) =lim
?????? →�
+
�(??????) =lim
?????? →�
−
�(??????) = ??????
When we make x approaches to a coming from the right, we use values
that are greater than a but close to the value of a. we call this, the right-hand
side of f as x approaches to a.
1. Evaluate lim
??????→5
−
√5− ?????? and lim
??????→5
+
√5−??????
Solution:
When the values are coming from the right side of 5, f(x) = √5− ?????? is not
defined. Therefore, lim
??????→5
+
√5−?????? does not exist.
When the values are coming from the left side of 5, f(x) = √5− ?????? is
defined. Therefore, lim
??????→5
−
√5−?????? = √5−4 = 1
1. Consider the function f defined by f(x) =
3,??????>1
1,??????=1
−2??????<1
. Evaluate each of the
following limits, if it exists.
a. lim
??????→1
+
�(??????) b. lim
??????→1
−
�(??????) c. lim
??????→1
�(??????)
Example 10
Example 11
Definition of Right-hand side Limit.
Let f be a function defined at every number in some open interval (a,d). Then
the limit of f(x) as x approaches a from the left side L written as
lim
?????? →�
+
�(??????) = ??????
Theorem 9. Existence of Limit
The limit of f as x approaches a exists and is equal to the number L if and only if
lim
?????? →�
−
�(??????) ��� lim
??????→�
+
�(??????) both exist and are equal to L. That is,
lim
??????→�
�(??????) =lim
?????? →�
+
�(??????) =lim
?????? →�
−
�(??????) = ??????
27
Solution:
a. To evaluate lim
??????→1
+
�(??????): Consider the values when x > 1, so we will use
f(x) = 3.
lim
??????→1
+
�(??????) = lim
??????→1
+
3 = 3
b. To evaluate lim
??????→1
−
�(??????): Consider the values when x < 1, so we will use
f(x) = -2.
lim
??????→1
−
�(??????) = lim
??????→1
−
−2 = -2
c. Therefore, lim
??????→1
�(??????) does not exist since lim
??????→1
+
�(??????) ≠lim
??????→1
−
�(??????).
Consider the function f defined as f(x) =
−??????
2
+ 4 ??????=−3
1 ??????=3
Evaluate each of the following limits if it exists.
a. lim
??????→−3
+
�(??????) b. lim
??????→−3
−
�(??????) c. lim
??????→−3
�(??????)
Solution:
Note that x ≠ 3 means that x > -3 or x <-3.
a. When evaluating the right-hand limit, let us get the values close to -3, that is,
the values greater than -3. Use the definition, -x
2
+ 4. Thus, lim
??????→−3
+
(−??????
2
+ 4) =
-5.
b. Similarly, we will use the definition f(x)= -x
2
+ 4 when evaluating the left-
hand limit.
Thus, lim
??????→−3
−
(−??????
2
+ 4) = -5.
c. Since lim
??????→−3
+
(−??????
2
+ 4) = -5= lim
??????→−3
−
(−??????
2
+ 4) = -5,
then lim
??????→−3
(−??????
2
+ 4) = -5.
Example 12
Solution:
a. To evaluate lim
??????→1
+
�(??????): Consider the values when x > 1, so we will use
f(x) = 3.
lim
??????→1
+
�(??????) = lim
??????→1
+
3 = 3
b. To evaluate lim
??????→1
−
�(??????): Consider the values when x < 1, so we will use
f(x) = -2.
lim
??????→1
−
�(??????) = lim
??????→1
−
−2 = -2
c. Therefore, lim
??????→1
�(??????) does not exist since lim
??????→1
+
�(??????) ≠lim
??????→1
−
�(??????).
Consider the function f defined as f(x) =
−??????
2
+ 4 ??????=−3
1 ??????=3
Evaluate each of the following limits if it exists.
a. lim
??????→−3
+
�(??????) b. lim
??????→−3
−
�(??????) c. lim
??????→−3
�(??????)
Solution:
Note that x ≠ 3 means that x > -3 or x <-3.
a. When evaluating the right-hand limit, let us get the values close to -3, that is,
the values greater than -3. Use the definition, -x
2
+ 4. Thus, lim
??????→−3
+
(−??????
2
+ 4) =
-5.
b. Similarly, we will use the definition f(x)= -x
2
+ 4 when evaluating the left-
hand limit.
Thus, lim
??????→−3
−
(−??????
2
+ 4) = -5.
c. Since lim
??????→−3
+
(−??????
2
+ 4) = -5= lim
??????→−3
−
(−??????
2
+ 4) = -5,
then lim
??????→−3
(−??????
2
+ 4) = -5.
Example 12
28
We have already considered the case when evaluating the limit of
rational function that leads to lim
??????→�
�(??????) = lim
??????→�
�(??????)
�(??????)
=
0
0
. In order to evaluate the
limit, we simplified such function by means of factoring and cancelling common
factors.
Now, we will consider cases wherein evaluating lim
??????→�
�(??????) = lim
??????→�
�(??????)
�(??????)
gives
a nonzero constant for p(a) while q(a) =0.
Consider the graph of function f(x)=
1
??????
in figure 1.
Figure 3
Suppose we want to evaluate lim
?????? →0
−
1
??????
and lim
?????? →0
+
1
??????
. Based from the graph,
as x approaches 0 from the left, the graph f tends to decrease infinitely without
bound. To represent the undefined extremely small value, we write the negative
infinity symbol as -∞. Now, as x approaches 0 from the right, the graph tends to
increase infinitely without bound. We write this positive infinity as +∞ or
simply ∞.
Self-Assessment Questions 1.5
2.6 INFINITE LIMITS
Consider the function f defined by f(x) =
2??????>1
1??????=1 .
−????????????<1
Evaluate each of the following
limits, if it exists:
1. lim
??????→1
+
�(??????)
2. lim
??????→1
−
�(??????)
3. lim
??????→1
�(??????)
We have already considered the case when evaluating the limit of
rational function that leads to lim
??????→�
�(??????) = lim
??????→�
�(??????)
�(??????)
=
0
0
. In order to evaluate the
limit, we simplified such function by means of factoring and cancelling common
factors.
Now, we will consider cases wherein evaluating lim
??????→�
�(??????) = lim
??????→�
�(??????)
�(??????)
gives
a nonzero constant for p(a) while q(a) =0.
Consider the graph of function f(x)=
1
??????
in figure 1.
Figure 3
Suppose we want to evaluate lim
?????? →0
−
1
??????
and lim
?????? →0
+
1
??????
. Based from the graph,
as x approaches 0 from the left, the graph f tends to decrease infinitely without
bound. To represent the undefined extremely small value, we write the negative
infinity symbol as -∞. Now, as x approaches 0 from the right, the graph tends to
increase infinitely without bound. We write this positive infinity as +∞ or
simply ∞.
Self-Assessment Questions 1.5
2.6 INFINITE LIMITS
Consider the function f defined by f(x) =
2??????>1
1??????=1 .
−????????????<1
Evaluate each of the following
limits, if it exists:
1. lim
??????→1
+
�(??????)
2. lim
??????→1
−
�(??????)
3. lim
??????→1
�(??????)
29
Now, we can also verify this example using a tabular approach. As we
can see below, as x approaches 0 from the left, f decreases infinitely and as x
approaches 0 from the right, f tends to increases infinitely.
x f(x)=
1
??????
x f(x)=
1
??????
-1 -1 1 1
-0.8 -1.25 0.8 1.25
-0.5 -2 0.5 2
-0.1 -10 0.1 10
-0.01 -100 0.01 100
-0.001 -1000 0.001 1000
-0.0001 -10000 0.0001 10000
-0.00001 -100000 0.00001 100000
-0.000001 -1000000 0.000001 1000000
Therefore, we can say that
lim
?????? →�
−
1
??????
= -∞ and lim
?????? →�
+
1
??????
= +∞ . Using the theorem 9 on existence of
limits,
lim
?????? →�
−
1
??????
= -∞ ≠ lim
?????? →�
+
1
??????
= +∞,lim
?????? →�
1
??????
does not exist.
The use of tabular approach in evaluating infinite limits can show the
values if it is increasing or decreasing infinitely. However, there is another way
to evaluate infinite limits without using tabular approach. Applying theorem 10
and by using the properties of infinity, we can evaluate infinite limits easily.
Theorem 10 tell us the sign of the function if the numerator is positive
but the denominator can be either positive or negative. The sign of the infinity
depends on the rule of number signs on division.
Theorem 10. Infinite Limits
Let r be a positive integer
1. If r is even, then
a. lim
?????? →0
−
1
??????
??????
= +∞ b. lim
?????? →0
+
1
??????
??????
= +∞
2. If r is odd, then
a. lim
?????? →0
−
1
??????
??????
= -∞ b. lim
?????? →0
+
1
??????
??????
= +∞
c. lim
?????? →0
−
1
√??????
?????? = -∞ d. lim
?????? →0
+
1
√??????
?????? = +∞
Now, we can also verify this example using a tabular approach. As we
can see below, as x approaches 0 from the left, f decreases infinitely and as x
approaches 0 from the right, f tends to increases infinitely.
x f(x)=
1
??????
x f(x)=
1
??????
-1 -1 1 1
-0.8 -1.25 0.8 1.25
-0.5 -2 0.5 2
-0.1 -10 0.1 10
-0.01 -100 0.01 100
-0.001 -1000 0.001 1000
-0.0001 -10000 0.0001 10000
-0.00001 -100000 0.00001 100000
-0.000001 -1000000 0.000001 1000000
Therefore, we can say that
lim
?????? →�
−
1
??????
= -∞ and lim
?????? →�
+
1
??????
= +∞ . Using the theorem 9 on existence of
limits,
lim
?????? →�
−
1
??????
= -∞ ≠ lim
?????? →�
+
1
??????
= +∞,lim
?????? →�
1
??????
does not exist.
The use of tabular approach in evaluating infinite limits can show the
values if it is increasing or decreasing infinitely. However, there is another way
to evaluate infinite limits without using tabular approach. Applying theorem 10
and by using the properties of infinity, we can evaluate infinite limits easily.
Theorem 10 tell us the sign of the function if the numerator is positive
but the denominator can be either positive or negative. The sign of the infinity
depends on the rule of number signs on division.
Theorem 10. Infinite Limits
Let r be a positive integer
1. If r is even, then
a. lim
?????? →0
−
1
??????
??????
= +∞ b. lim
?????? →0
+
1
??????
??????
= +∞
2. If r is odd, then
a. lim
?????? →0
−
1
??????
??????
= -∞ b. lim
?????? →0
+
1
??????
??????
= +∞
c. lim
?????? →0
−
1
√??????
?????? = -∞ d. lim
?????? →0
+
1
√??????
?????? = +∞
30
1. Evaluate each of the following functions if it exists:
a. lim
??????→0
+
1
??????
7
b. lim
??????→0
+
1
??????
4
c. lim
??????→0
−
1
??????
11
Solution:
a. lim
??????→0
+
1
??????
7
= +∞
b. lim
??????→0
+
1
??????
4
= +∞
c. lim
??????→0
−
1
??????
11
= -∞
1. Evaluate lim
??????→ −3
−
(
1
3 + ??????
+
??????−1
??????
2
)
Solution:
lim
??????→−3
−
1
3 + ??????
=
1
3 + −3
=-∞
lim
??????→−3
−
??????−1
??????
2
= −
4
9
= -
2
3
Hence, by Theorem 11a, lim
??????→ −3
−
(
1
3 + ??????
+
??????−1
??????
2
) = lim
??????→ −3
−
(−∞ +
−2
3
)= -∞
Example 12
Self-Assessment Questions 1.6
Example 13
Example 14
Theorem 11a. Properties of Infinite Limits (Sum)
If lim
??????→�
�(??????) = ±∞ and lim
??????→�
�(??????) = �, then lim
??????→�
[�(??????) + �(??????)] = ±∞
Theorem 11b. Properties of Infinite Limits (Product)
Let lim
??????→�
�(??????) = �≠0,
i. lim
??????→�
[�(??????) ∙ �(??????)] = +∞ if lim
??????→�
�(??????) = +∞ and c > 0, or lim
??????→�
�(??????) = −∞ and c < 0
ii. lim
??????→�
[�(??????) ∙ �(??????)] = −∞ if lim
??????→�
�(??????) = −∞ and c >0, or lim
??????→�
�(??????) = +∞ and c < 0
1. Evaluate each of the following functions if it exists:
a. lim
??????→0
+
1
??????
7
b. lim
??????→0
+
1
??????
4
c. lim
??????→0
−
1
??????
11
Solution:
a. lim
??????→0
+
1
??????
7
= +∞
b. lim
??????→0
+
1
??????
4
= +∞
c. lim
??????→0
−
1
??????
11
= -∞
1. Evaluate lim
??????→ −3
−
(
1
3 + ??????
+
??????−1
??????
2
)
Solution:
lim
??????→−3
−
1
3 + ??????
=
1
3 + −3
=-∞
lim
??????→−3
−
??????−1
??????
2
= −
4
9
= -
2
3
Hence, by Theorem 11a, lim
??????→ −3
−
(
1
3 + ??????
+
??????−1
??????
2
) = lim
??????→ −3
−
(−∞ +
−2
3
)= -∞
Example 12
Self-Assessment Questions 1.6
Example 13
Example 14
Theorem 11a. Properties of Infinite Limits (Sum)
If lim
??????→�
�(??????) = ±∞ and lim
??????→�
�(??????) = �, then lim
??????→�
[�(??????) + �(??????)] = ±∞
Theorem 11b. Properties of Infinite Limits (Product)
Let lim
??????→�
�(??????) = �≠0,
i. lim
??????→�
[�(??????) ∙ �(??????)] = +∞ if lim
??????→�
�(??????) = +∞ and c > 0, or lim
??????→�
�(??????) = −∞ and c < 0
ii. lim
??????→�
[�(??????) ∙ �(??????)] = −∞ if lim
??????→�
�(??????) = −∞ and c >0, or lim
??????→�
�(??????) = +∞ and c < 0
31
1. lim
??????→0
+
(
3
??????
2
)(?????? −
1
3
)
Solution:
lim
??????→0
+
(
3
??????
2
) = +∞
lim
??????→0
+
(?????? −
1
3
) = 0-
1
3
=
−1
3
Hence, by Theorem 11b, lim
??????→0
+
(
3
??????
2
)(?????? −
1
3
)= -∞
Note that “g(x) →0
+
“ and “ g(x) →0
−
“ mean that the values of g become closer
and closer to 0 through positive and negative values respectively.
1. Evaluate lim
??????→−3
−
??????−4
??????
2
− 9
Solution:
lim
??????→−3
−
??????−4 = -3 -4 = -7
lim
??????→−3
−
(??????
2
− 9) = 0
Since lim
??????→−3
−
??????−4= −7, c<0 and lim
??????→−3
−
(??????
2
−9) = 0. To determine
whether x
2
– 9 is approaching 0 through positive or negative values, choose a
value of x that is close to, but less that of -3, say x = -3.01 and put it into x
2
– 9.
Observe that x
2
– 9 = (-3.01)
2
– 9 = 0.06 < 0. Thus, (x
2
– 9) → 0
+
,
Hence, by Theorem 11c, lim
??????→−3
−
??????−4
??????
2
− 9
= -∞
Example 15
Theorem 11c. Properties of Infinite Limits (Quotient)
Let lim
??????→�
�(??????) = �≠0, and lim
??????→�
�(??????) =0, then
i. lim
??????→�
�(??????)
�(??????)
= +∞ if c > 0 and g(x) → 0
+
, or if c < 0 and g(x) → 0
−
ii. lim
??????→�
�(??????)
�(??????)
= -∞ if c > 0 and g(x) → 0
−
,or if c < 0 and g(x) → 0
+
1. lim
??????→0
+
(
3
??????
2
)(?????? −
1
3
)
Solution:
lim
??????→0
+
(
3
??????
2
) = +∞
lim
??????→0
+
(?????? −
1
3
) = 0-
1
3
=
−1
3
Hence, by Theorem 11b, lim
??????→0
+
(
3
??????
2
)(?????? −
1
3
)= -∞
Note that “g(x) →0
+
“ and “ g(x) →0
−
“ mean that the values of g become closer
and closer to 0 through positive and negative values respectively.
1. Evaluate lim
??????→−3
−
??????−4
??????
2
− 9
Solution:
lim
??????→−3
−
??????−4 = -3 -4 = -7
lim
??????→−3
−
(??????
2
− 9) = 0
Since lim
??????→−3
−
??????−4= −7, c<0 and lim
??????→−3
−
(??????
2
−9) = 0. To determine
whether x
2
– 9 is approaching 0 through positive or negative values, choose a
value of x that is close to, but less that of -3, say x = -3.01 and put it into x
2
– 9.
Observe that x
2
– 9 = (-3.01)
2
– 9 = 0.06 < 0. Thus, (x
2
– 9) → 0
+
,
Hence, by Theorem 11c, lim
??????→−3
−
??????−4
??????
2
− 9
= -∞
Example 15
Theorem 11c. Properties of Infinite Limits (Quotient)
Let lim
??????→�
�(??????) = �≠0, and lim
??????→�
�(??????) =0, then
i. lim
??????→�
�(??????)
�(??????)
= +∞ if c > 0 and g(x) → 0
+
, or if c < 0 and g(x) → 0
−
ii. lim
??????→�
�(??????)
�(??????)
= -∞ if c > 0 and g(x) → 0
−
,or if c < 0 and g(x) → 0
+
32
1. Evaluate lim
??????→5
+
2??????+1
??????− 5
Solution:
lim
??????→5
+
2?????? + 1 = 11 > 0
lim
??????→5
+
??????−5 = 0
To determine whether x -5 is approaching 0 through positive or negative
values, choose a value of x that is close to, but greater than 5, say 5.01, and put
it into x -5. Observe that x – 5 = 5.01 – 5 = 0.01 > 0. Thus, (x-5) → 0
+
. Therefore,
lim
??????→5
+
2??????+1
??????− 5
= +∞
The limit of 1x as x approaches Infinity is 0
And write it like this:
In this section, we are going to observe the behavior of a function as the
independent variable increases or decreases without bound.
We consider the function f(x) =
1
??????
. We evaluate the given function using
numbers that become extremely small and extremely large. Notice in table 1.7.1
that as x becomes larger and larger, the value of f becomes closer to 0. The limit
of f(x) as x increases without bound is 0. In symbol,
lim
??????→+∞
1
??????
= 0
Example 16
Self-Assessment Questions 1.6
2.7 LIMITS AT INFINITY
Evaluate the following limits, if it exists:
1.lim
??????→−1
3
?????? + 2
2. lim
??????→−5
−
(
3
?????? + 5
+
??????−1
??????
2
)
3. lim
??????→4
−
(
3
(??????−4)
2
)(
??????
2
−2
5??????
)
1. Evaluate lim
??????→5
+
2??????+1
??????− 5
Solution:
lim
??????→5
+
2?????? + 1 = 11 > 0
lim
??????→5
+
??????−5 = 0
To determine whether x -5 is approaching 0 through positive or negative
values, choose a value of x that is close to, but greater than 5, say 5.01, and put
it into x -5. Observe that x – 5 = 5.01 – 5 = 0.01 > 0. Thus, (x-5) → 0
+
. Therefore,
lim
??????→5
+
2??????+1
??????− 5
= +∞
The limit of 1x as x approaches Infinity is 0
And write it like this:
In this section, we are going to observe the behavior of a function as the
independent variable increases or decreases without bound.
We consider the function f(x) =
1
??????
. We evaluate the given function using
numbers that become extremely small and extremely large. Notice in table 1.7.1
that as x becomes larger and larger, the value of f becomes closer to 0. The limit
of f(x) as x increases without bound is 0. In symbol,
lim
??????→+∞
1
??????
= 0
Example 16
Self-Assessment Questions 1.6
2.7 LIMITS AT INFINITY
Evaluate the following limits, if it exists:
1.lim
??????→−1
3
?????? + 2
2. lim
??????→−5
−
(
3
?????? + 5
+
??????−1
??????
2
)
3. lim
??????→4
−
(
3
(??????−4)
2
)(
??????
2
−2
5??????
)
33
x f(x)
1 1
10 0.1
100 0.01
1,000 0.001
10,000 0.0001
100,000 0.00001
1,000,000 0.000001
Table 1.7.1
Now observe what happens to the function f(x) as x becomes smaller
and smaller. In table 1.7.2, we can see that the values become closer to 0 as x
becomes smaller and smaller. We say that the limit of f(x) as x decreases
without bound is 0 and we write this in symbol,
lim
??????→−∞
1
??????
= 0
X f(x)
1 1
10 0.1
100 0.01
1,000 0.001
10,000 0.0001
100,000 0.00001
1,000,000 0.000001
Table 1.7.2
In other words:
Definition 1.7.1
Let f be a function defined on (c, +∞) for some number c. The limit of f as
increases without bound L, which is written as
lim
??????→+∞
�(??????) =L
if the values of f(x) get closer to one and only one number L as x becomes
larger and larger.
Definition 1.7.2
Let f be a function defined on (c, +∞) for some number c. The limit of f as decreases
without bound L, which is written as
lim
??????→−∞
�(??????) =L
if the values of f(x) get closer to one and only one number L as x becomes
smaller and smaller.
x f(x)
1 1
10 0.1
100 0.01
1,000 0.001
10,000 0.0001
100,000 0.00001
1,000,000 0.000001
Table 1.7.1
Now observe what happens to the function f(x) as x becomes smaller
and smaller. In table 1.7.2, we can see that the values become closer to 0 as x
becomes smaller and smaller. We say that the limit of f(x) as x decreases
without bound is 0 and we write this in symbol,
lim
??????→−∞
1
??????
= 0
X f(x)
1 1
10 0.1
100 0.01
1,000 0.001
10,000 0.0001
100,000 0.00001
1,000,000 0.000001
Table 1.7.2
In other words:
Definition 1.7.1
Let f be a function defined on (c, +∞) for some number c. The limit of f as
increases without bound L, which is written as
lim
??????→+∞
�(??????) =L
if the values of f(x) get closer to one and only one number L as x becomes
larger and larger.
Definition 1.7.2
Let f be a function defined on (c, +∞) for some number c. The limit of f as decreases
without bound L, which is written as
lim
??????→−∞
�(??????) =L
if the values of f(x) get closer to one and only one number L as x becomes
smaller and smaller.
34
The limits:lim
??????→+∞
�(??????) =0 and lim
??????→−∞
�(??????) =0 is what we call limits at
infinity. Now to generalize results and evaluation of these limits, we have the
following theorem:
1. Evaluate the indicated limits:
a. lim
??????→+∞
4??????
2
+2??????−1
2??????
2
+ 3??????+1
Solution:
To apply Theorem 1.7.1, we must each term of the polynomial in the
numerator and denominator in the form of
??????
??????
??????
. to do this, we divide each term,
both in the numerator and denominator by x
2
. Dividing both the numerator and
denominator by x
2
does not change the given.
lim
??????→+∞
4??????
2
+2??????−1
2??????
2
+ 3??????+1
= lim
??????→+∞
4??????
2
??????
2
+
2??????
??????
2
−
1
??????
2
2??????
2
??????
2
+
3??????
??????
2
+
1
??????
2
= lim
??????→+∞
4 +
2
??????
−
1
??????
2
2 +
3
??????
+
1
??????
2
Apply limits laws and theorem 1.7.1. Then,
lim
??????→+∞
4 +
2
??????
−
1
??????
2
2 +
3
??????
+
1
??????
2
=
lim
??????→+∞
4 + lim
??????→+∞
2
??????
− lim
??????→+∞
1
??????
2
lim
??????→+∞
2 +lim
??????→+∞
3
??????
+ lim
??????→+∞
1
??????
2
=
4 + 0 − 0
2 + 0 + 0
= 2
Example 17
Theorem 1.7.3
Let k be any real number and r be any positive rational number, then
i. lim
?????? →+∞
??????
??????
??????
= 0
ii. lim
?????? →−∞
??????
??????
??????
= 0, provided x
r is defined when using x< 0.
The limits:lim
??????→+∞
�(??????) =0 and lim
??????→−∞
�(??????) =0 is what we call limits at
infinity. Now to generalize results and evaluation of these limits, we have the
following theorem:
1. Evaluate the indicated limits:
a. lim
??????→+∞
4??????
2
+2??????−1
2??????
2
+ 3??????+1
Solution:
To apply Theorem 1.7.1, we must each term of the polynomial in the
numerator and denominator in the form of
??????
??????
??????
. to do this, we divide each term,
both in the numerator and denominator by x
2
. Dividing both the numerator and
denominator by x
2
does not change the given.
lim
??????→+∞
4??????
2
+2??????−1
2??????
2
+ 3??????+1
= lim
??????→+∞
4??????
2
??????
2
+
2??????
??????
2
−
1
??????
2
2??????
2
??????
2
+
3??????
??????
2
+
1
??????
2
= lim
??????→+∞
4 +
2
??????
−
1
??????
2
2 +
3
??????
+
1
??????
2
Apply limits laws and theorem 1.7.1. Then,
lim
??????→+∞
4 +
2
??????
−
1
??????
2
2 +
3
??????
+
1
??????
2
=
lim
??????→+∞
4 + lim
??????→+∞
2
??????
− lim
??????→+∞
1
??????
2
lim
??????→+∞
2 +lim
??????→+∞
3
??????
+ lim
??????→+∞
1
??????
2
=
4 + 0 − 0
2 + 0 + 0
= 2
Example 17
Theorem 1.7.3
Let k be any real number and r be any positive rational number, then
i. lim
?????? →+∞
??????
??????
??????
= 0
ii. lim
?????? →−∞
??????
??????
??????
= 0, provided x
r is defined when using x< 0.
35
1. Evaluate limits:
lim 4x
3
+ 3x
2
- 6
x→∞ 2x
3
+ 5x + 3
Solution:
The function assumes the indeterminate form
∞
∞
. Dividing the numerator
and denominator by x
3
, we get
lim 4 + 3/x – 6/x
3
= 4 + 0 – 0 = 2
x→∞ 2 + 5/x
2
+ 3/x
3
2 + 0 – 0
1. Evaluate limits:
lim x
2
+ 2x – 6 =
x→-∞ x
3
+ 4
Solution:
The function assumes the indeterminate form
∞
∞
. Dividing the numerator
and denominator by x
3
, we get
lim x
2
+ 2x – 6 =
x→-∞ x
3
+ 4
= lim x
2
/x
3
+ 2x /x
3
– 6/x
3
x→-∞ x
3
/x
3
+ 4 /x
3
=
lim
??????→−∞
1
??????
+ lim
??????→−∞
2
??????
2
− lim
??????→−∞
6
??????
3
lim
??????→−∞
1 + lim
??????→−∞
4
??????
3
=
0 + 0 − 0
1 + 0
= 0
Example 18
Example 19
1. Evaluate limits:
lim 4x
3
+ 3x
2
- 6
x→∞ 2x
3
+ 5x + 3
Solution:
The function assumes the indeterminate form
∞
∞
. Dividing the numerator
and denominator by x
3
, we get
lim 4 + 3/x – 6/x
3
= 4 + 0 – 0 = 2
x→∞ 2 + 5/x
2
+ 3/x
3
2 + 0 – 0
1. Evaluate limits:
lim x
2
+ 2x – 6 =
x→-∞ x
3
+ 4
Solution:
The function assumes the indeterminate form
∞
∞
. Dividing the numerator
and denominator by x
3
, we get
lim x
2
+ 2x – 6 =
x→-∞ x
3
+ 4
= lim x
2
/x
3
+ 2x /x
3
– 6/x
3
x→-∞ x
3
/x
3
+ 4 /x
3
=
lim
??????→−∞
1
??????
+ lim
??????→−∞
2
??????
2
− lim
??????→−∞
6
??????
3
lim
??????→−∞
1 + lim
??????→−∞
4
??????
3
=
0 + 0 − 0
1 + 0
= 0
Example 18
Example 19
36
Continuity of a function describes its graph whether it contains broken
part at a given number or interval. When it is unbroken at a number or interval,
the function is said to be continuous. This means that it has no holes or gaps.
Consider a rational function f(x) =
??????
2
+ ??????−1
??????
2
−1
whose domain is the set of real
numbers except -1 and 1. This function is undefined at -1 and 1. We say that the
function is not continuous at these numbers or is said to be discontinuous there.
The graph shown below is an example of discontinuous function.
Figure 4
Graph of
??????
4
−??????
3
+??????−1
??????−1
Self-Assessment Questions 1.7
2.8 CONTINUITY OF A FUNCTION
Evaluate the following limits, if it exists:
1. lim 6x
3 + 4x
2 + 5
x→∞ 8x
3 + 7x - 3
2. lim 4x + 5
x→∞ x
2 + 1
3. lim x
4
+ 1
x → -∞ x
3
+ 3x
This graph was drawn
through
https://www.geogeb
ra.org/graphing?lang
=en
Continuity of a function describes its graph whether it contains broken
part at a given number or interval. When it is unbroken at a number or interval,
the function is said to be continuous. This means that it has no holes or gaps.
Consider a rational function f(x) =
??????
2
+ ??????−1
??????
2
−1
whose domain is the set of real
numbers except -1 and 1. This function is undefined at -1 and 1. We say that the
function is not continuous at these numbers or is said to be discontinuous there.
The graph shown below is an example of discontinuous function.
Figure 4
Graph of
??????
4
−??????
3
+??????−1
??????−1
Self-Assessment Questions 1.7
2.8 CONTINUITY OF A FUNCTION
Evaluate the following limits, if it exists:
1. lim 6x
3 + 4x
2 + 5
x→∞ 8x
3 + 7x - 3
2. lim 4x + 5
x→∞ x
2 + 1
3. lim x
4
+ 1
x → -∞ x
3
+ 3x
This graph was drawn
through
https://www.geogeb
ra.org/graphing?lang
=en
37
The figures below show graphical examples of functions where either
1, 2 or 3 can fail to hold.
Figure 5
Definition 1.8.1 Continuous Function.
A function f is continuous at a point if and only if lim
??????→�
�(??????)= �(�).
The definition implies that three condition must be satisfied for a function f to be
continuous at a. These are the following:
i. f(a) exists;
ii. lim
??????→�
�(??????) exists;
iii. lim
??????→�
�(??????)= �(�).
The figures below show graphical examples of functions where either
1, 2 or 3 can fail to hold.
Figure 5
Definition 1.8.1 Continuous Function.
A function f is continuous at a point if and only if lim
??????→�
�(??????)= �(�).
The definition implies that three condition must be satisfied for a function f to be
continuous at a. These are the following:
i. f(a) exists;
ii. lim
??????→�
�(??????) exists;
iii. lim
??????→�
�(??????)= �(�).
38
If f is not able to satisfy at least one of these three conditions, we say
that f has a discontinuity at a or f is discontinuous at a.
Types of Discontinuity
1. If f is discontinuous at a and condition (ii) is not satisfied, the discontinuity
is called essential.
2. If f satisfies condition (ii) but fails conditions (i) or (iii), the discontinuity is
called removable. A function that has a removable discontinuity at a point can
be redefined or manipulated straightforwardly so that it will no longer be
discontinuous at that point.
Determine if the function is continuous at the given number a. If the
function is discontinuous at a, determine the type of discontinuity involved.
1. f(x) =
−1 ??????� ?????? ≤ 2
?????? =−3??????� ?????? > 2
a= 2
Solution:
The three conditions for a continuity of function must be satisfied.
1. f(x) =
−1??????� ?????? ≤ 2
??????−3??????� ?????? > 2
a= 2
i. f(2) = -1
ii. For lim
??????→2
�(??????), we need to check the left-hand
and right-hand limits and compare if they are equal.
lim
??????→2
+
�(??????) = lim
??????→2
+
(−1) = -1
lim
??????→2
−
�(??????) = lim
??????→2
−
(??????−3)= 2 -3 = -1
Figure 6
iii. lim
??????→2
�(??????) = f(2)
Since all the three conditions are satisfied, f is continuous at 2.
Example 20
If f is not able to satisfy at least one of these three conditions, we say
that f has a discontinuity at a or f is discontinuous at a.
Types of Discontinuity
1. If f is discontinuous at a and condition (ii) is not satisfied, the discontinuity
is called essential.
2. If f satisfies condition (ii) but fails conditions (i) or (iii), the discontinuity is
called removable. A function that has a removable discontinuity at a point can
be redefined or manipulated straightforwardly so that it will no longer be
discontinuous at that point.
Determine if the function is continuous at the given number a. If the
function is discontinuous at a, determine the type of discontinuity involved.
1. f(x) =
−1 ??????� ?????? ≤ 2
?????? =−3??????� ?????? > 2
a= 2
Solution:
The three conditions for a continuity of function must be satisfied.
1. f(x) =
−1??????� ?????? ≤ 2
??????−3??????� ?????? > 2
a= 2
i. f(2) = -1
ii. For lim
??????→2
�(??????), we need to check the left-hand
and right-hand limits and compare if they are equal.
lim
??????→2
+
�(??????) = lim
??????→2
+
(−1) = -1
lim
??????→2
−
�(??????) = lim
??????→2
−
(??????−3)= 2 -3 = -1
Figure 6
iii. lim
??????→2
�(??????) = f(2)
Since all the three conditions are satisfied, f is continuous at 2.
Example 20
39
Determine if the function is continuous at the given number a. If the function is
discontinuous at a, determine the type of discontinuity involved.
f(x) =
2??????−3
??????+ 3
a= -3
i. When x =-3, the value of f(-3) is not defined which
makes condition (i) not satisfied.
So this function has an infinite
discontinuity at -3 as
shown in the graph.
Then x has a removable discontinuity.
Figure 7
Determine if the function is continuous at the given number a. If the function is
discontinuous at a, determine the type of discontinuity involved.
1. f(x) =
√??????−2??????≥2
??????
3
−1??????<2
; a=2
Solution:
i. f(2) = √??????−2 = √2−2 = 0
ii. lim
??????→2
+
√??????−2 = lim
??????→2
+
√2−2 = 0
lim
??????→2
−
??????
3
−1 = lim
??????→2
−
2
3
−1 = 7
Since lim
??????→2
+
�(??????)≠ lim
??????→2
−
�(??????), then lim
??????→2
�(??????) does not exist. This means that x
has an essential discontinuity at 2.
1. Evaluate the function f(x) = x
2
if it is continuous when x = 2
Solution:
Since lim x
2
= f(2) = 4, then the function f(x) = x
2
is continuous at x = 2.
x→2
Example 21
Example 22
Example 23
Determine if the function is continuous at the given number a. If the function is
discontinuous at a, determine the type of discontinuity involved.
f(x) =
2??????−3
??????+ 3
a= -3
i. When x =-3, the value of f(-3) is not defined which
makes condition (i) not satisfied.
So this function has an infinite
discontinuity at -3 as
shown in the graph.
Then x has a removable discontinuity.
Figure 7
Determine if the function is continuous at the given number a. If the function is
discontinuous at a, determine the type of discontinuity involved.
1. f(x) =
√??????−2??????≥2
??????
3
−1??????<2
; a=2
Solution:
i. f(2) = √??????−2 = √2−2 = 0
ii. lim
??????→2
+
√??????−2 = lim
??????→2
+
√2−2 = 0
lim
??????→2
−
??????
3
−1 = lim
??????→2
−
2
3
−1 = 7
Since lim
??????→2
+
�(??????)≠ lim
??????→2
−
�(??????), then lim
??????→2
�(??????) does not exist. This means that x
has an essential discontinuity at 2.
1. Evaluate the function f(x) = x
2
if it is continuous when x = 2
Solution:
Since lim x
2
= f(2) = 4, then the function f(x) = x
2
is continuous at x = 2.
x→2
Example 21
Example 22
Example 23
40
The graph of the function is shown:
Figure 8
1. Determine if f(x) =
??????
2
− ?????? − 2
??????−2
is continuous or not at x=0
Solution:
We have to check the three conditions for continuity of a function
i. if x = 0, then f(0) = 1
ii. lim
??????→0
�(??????)= lim
??????→0
??????
2
− ?????? − 2
??????−2
= lim
??????→0
0
2
− 0 − 2
0−2
= 1
iii. f(0) = 1 = lim
??????→0
�(??????)
The limit of 1x as x approaches Infinity is 0
And write it like this:
Example 24
Self-Assessment Questions 1.8
Determine if the function is continuous at the given number a. If the function is
discontinuous at a, state which of the conditions is/are not satisfied and
determine the type of discontinuity involved.
1. f(x) =
??????
2
+ 1
−?????? + 1
; a = 2
2. g(x) =
?????? + 2,??????<0
−1, ??????=0
−??????
2
+ 2,?????? >0
; a= 0
3. h(x) =
1−??????,?????? <−1
??????
2
−1,??????≥ −1
; a = -1
The graph of the function is shown:
Figure 8
1. Determine if f(x) =
??????
2
− ?????? − 2
??????−2
is continuous or not at x=0
Solution:
We have to check the three conditions for continuity of a function
i. if x = 0, then f(0) = 1
ii. lim
??????→0
�(??????)= lim
??????→0
??????
2
− ?????? − 2
??????−2
= lim
??????→0
0
2
− 0 − 2
0−2
= 1
iii. f(0) = 1 = lim
??????→0
�(??????)
The limit of 1x as x approaches Infinity is 0
And write it like this:
Example 24
Self-Assessment Questions 1.8
Determine if the function is continuous at the given number a. If the function is
discontinuous at a, state which of the conditions is/are not satisfied and
determine the type of discontinuity involved.
1. f(x) =
??????
2
+ 1
−?????? + 1
; a = 2
2. g(x) =
?????? + 2,??????<0
−1, ??????=0
−??????
2
+ 2,?????? >0
; a= 0
3. h(x) =
1−??????,?????? <−1
??????
2
−1,??????≥ −1
; a = -1
41
Given the work of the previous section, we are now in a position to state
a clear definition of the notion of continuity. We will have several related
definitions, but the fundamental definition is that of continuity at a point.
Intuitively, continuity at a point c for a function f means that the values of f for
points near a do not change abruptly from the value of f at a.
The three conditions of continuity at a number must be satisfied for each
number in the open interval. The three conditions can be summarized into one
condition, that is the condition(iii) which is lim
??????→�
�(??????) = �(�). This imply that it
is enough to check this condition to find the continuity at a number in an open
interval.
Polynomial functions are examples of functions which are always
continuous on any open interval. Other examples are rational functions like f(x)
=
??????
??????
2
+ 4
where the denominator cannot be zero.
If a function f is continuous on an open interval (a,b) and also continuous
from the right of a, and continuous from the left of b, then f is continuous on the
closed interval [a, b].
An interval is called open if it doesn’t include the endpoints. It is denoted
by ( ). For instance, (1, 2) means greater than 1 and less than 2.
Consider the function f(x) = 3x
3
− 6x + 1. Since f is a polynomial, it is
continuous on (−∞,∞). That is, for any real number c, f is continuous at c.
2.9 CONTINUITY ON AN INTERVAL
Example 25
Definition 1.9. Continuity on an Open Interval
A function f is continuous on an open interval (a,b) if f is continuous
at each number in the interval.
Definition 1.9. 1 Continuity on a Closed Interval
A function f is continuous on a closed interval [a, b] if
i. f is continuous on the open interval (a, b)
ii. lim
??????→�
+
�(??????)=�(�) and lim
??????→�
−
�(??????)=�(�)
Given the work of the previous section, we are now in a position to state
a clear definition of the notion of continuity. We will have several related
definitions, but the fundamental definition is that of continuity at a point.
Intuitively, continuity at a point c for a function f means that the values of f for
points near a do not change abruptly from the value of f at a.
The three conditions of continuity at a number must be satisfied for each
number in the open interval. The three conditions can be summarized into one
condition, that is the condition(iii) which is lim
??????→�
�(??????) = �(�). This imply that it
is enough to check this condition to find the continuity at a number in an open
interval.
Polynomial functions are examples of functions which are always
continuous on any open interval. Other examples are rational functions like f(x)
=
??????
??????
2
+ 4
where the denominator cannot be zero.
If a function f is continuous on an open interval (a,b) and also continuous
from the right of a, and continuous from the left of b, then f is continuous on the
closed interval [a, b].
An interval is called open if it doesn’t include the endpoints. It is denoted
by ( ). For instance, (1, 2) means greater than 1 and less than 2.
Consider the function f(x) = 3x
3
− 6x + 1. Since f is a polynomial, it is
continuous on (−∞,∞). That is, for any real number c, f is continuous at c.
2.9 CONTINUITY ON AN INTERVAL
Example 25
Definition 1.9. Continuity on an Open Interval
A function f is continuous on an open interval (a,b) if f is continuous
at each number in the interval.
Definition 1.9. 1 Continuity on a Closed Interval
A function f is continuous on a closed interval [a, b] if
i. f is continuous on the open interval (a, b)
ii. lim
??????→�
+
�(??????)=�(�) and lim
??????→�
−
�(??????)=�(�)
42
A closed interval is one which includes all the limit points. It is denoted
by [ ]. For example, [2, 5] means greater than or equal to 2 and less than or equal
to 5.
The function f(x) =
1
??????
2
−1
is not defined at -1 and 1. Thus, it is continuous on the
open interval (0,1). However, lim
??????→1
−
1
??????
2
−1
does not exist. Thus, it is not continuous
on [0, 1].
State the interval(s) over which a function f(x)=√4− ??????
2
is continuous.
Solution:
From the limit laws, we knew that the lim
??????→�
√4− ??????
2
= √4− �
2
for all
values of a in (-2,2). We also know that lim
??????→−2
+
√4− ??????
2
=0 exists and
lim
??????→2
−
√4− ??????
2
=0 exists. Therefore, f(x) is continuous over the interval [−2,2].
Determine if the function is continuous on the indicated closed interval.
f(x)= x√??????; [0,1]
Solution:
The domain of f(x)= x√?????? is the set of nonnegative numbers. Thus,
a. f(x) = x√?????? is continuous on the open
interval (0,1).
b. lim
??????→0
+
x√?????? = 0√0= 0
lim
??????→1
−
1√1 = 1√1= 1
Hence, f(x)= x√?????? is continuous at [0,1].
Figure 9
Example 26
Example 27
Example 28
A closed interval is one which includes all the limit points. It is denoted
by [ ]. For example, [2, 5] means greater than or equal to 2 and less than or equal
to 5.
The function f(x) =
1
??????
2
−1
is not defined at -1 and 1. Thus, it is continuous on the
open interval (0,1). However, lim
??????→1
−
1
??????
2
−1
does not exist. Thus, it is not continuous
on [0, 1].
State the interval(s) over which a function f(x)=√4− ??????
2
is continuous.
Solution:
From the limit laws, we knew that the lim
??????→�
√4− ??????
2
= √4− �
2
for all
values of a in (-2,2). We also know that lim
??????→−2
+
√4− ??????
2
=0 exists and
lim
??????→2
−
√4− ??????
2
=0 exists. Therefore, f(x) is continuous over the interval [−2,2].
Determine if the function is continuous on the indicated closed interval.
f(x)= x√??????; [0,1]
Solution:
The domain of f(x)= x√?????? is the set of nonnegative numbers. Thus,
a. f(x) = x√?????? is continuous on the open
interval (0,1).
b. lim
??????→0
+
x√?????? = 0√0= 0
lim
??????→1
−
1√1 = 1√1= 1
Hence, f(x)= x√?????? is continuous at [0,1].
Figure 9
Example 26
Example 27
Example 28
43
Feedback!
As you go along this module, what lesson did you feel the most difficult
for you? Try to list them down and give time to consult your teacher for further
discussion.
Self-Assessment Questions 1.9
Discuss the continuity of the function on the given interval.
1. f(x) = √25 −??????
2
; [-5,5]
2. f(x) =
??????
??????+ 3
; [-2,0]
Feedback!
As you go along this module, what lesson did you feel the most difficult
for you? Try to list them down and give time to consult your teacher for further
discussion.
Self-Assessment Questions 1.9
Discuss the continuity of the function on the given interval.
1. f(x) = √25 −??????
2
; [-5,5]
2. f(x) =
??????
??????+ 3
; [-2,0]
44
Exercise 2
Name: ____________________________________________ Score: ________________
Course and Year: _______________________________ Date : ________________
A. Evaluate the indicated limits using tabular approach.
1. lim
??????→5
(3?????? + 2)
B. Find the limits. Apply algebraic manipulation to evaluate the following
limits.
1. lim
??????→5
??????
2
− 25
??????
2
−5??????
2. lim
??????→10
10−??????
1
??????
−
1
10
3. lim
??????→1
??????
3
−3??????
2
Exercise 2
Name: ____________________________________________ Score: ________________
Course and Year: _______________________________ Date : ________________
A. Evaluate the indicated limits using tabular approach.
1. lim
??????→5
(3?????? + 2)
B. Find the limits. Apply algebraic manipulation to evaluate the following
limits.
1. lim
??????→5
??????
2
− 25
??????
2
−5??????
2. lim
??????→10
10−??????
1
??????
−
1
10
3. lim
??????→1
??????
3
−3??????
2
45
4. lim
??????→2
√5?????? + 6
5. lim
??????→1
??????
3
+ 2??????
2
− ??????−2
??????
3
+4??????
2
−5??????
C. Evaluate the following infinite limits and limits at infinity.
1. lim
??????→−5
+
−3
?????? + 5
2. lim
??????→1
−
6??????
??????
2
−1
3. lim
??????→−∞
(4??????
2
−3?????? + 5)
D. Determine if the function is continuous at the given number c. if it is
discontinuous at c, state the condition of continuity that is not satisfied.
1. f(x) =
??????
2
−4??????
??????
; c = 0
2. f(x) =
??????
2
??????� ?????? ≤2
?????? + 2??????� ??????>2
; c= 2
4. lim
??????→2
√5?????? + 6
5. lim
??????→1
??????
3
+ 2??????
2
− ??????−2
??????
3
+4??????
2
−5??????
C. Evaluate the following infinite limits and limits at infinity.
1. lim
??????→−5
+
−3
?????? + 5
2. lim
??????→1
−
6??????
??????
2
−1
3. lim
??????→−∞
(4??????
2
−3?????? + 5)
D. Determine if the function is continuous at the given number c. if it is
discontinuous at c, state the condition of continuity that is not satisfied.
1. f(x) =
??????
2
−4??????
??????
; c = 0
2. f(x) =
??????
2
??????� ?????? ≤2
?????? + 2??????� ??????>2
; c= 2
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