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7/25/2019 Chapter 1 Limits
1/23
CHAPTER
1Limits
Chapter Outline
1.1 Limits1.2 One Sided Limits
1.3 Limits Properties
1.4 Limits Involve Infinity
1.5 Continuity
Chapter Overview and Learning Objectives
This chapter presents to find limit of a function is an interesting concept where it may
be possible that value of the function does not exist at a point but we try to find the
value in the neighbourhood of the point. We will talk about this in more detail in the
chapter. In the other part of the chapter we will discuss continuity of a function which
is closely related to the concept of limits. There are some functions for which graph is
continuous while there are others for which this is not the case.
After careful study of this chapter, you should be able to do the following:
1. Clearly understand to identify the limits of a given function using the left and right
limits.
2. Clearly understand to define the continuity of a given function.
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1.1 Limits
A limit of a function )(xf at a number ax looks at the behavior of the
function )(xf as axgets closer and closer to the point ax as illustrated in Figure
1. In other word, we are looking at the value taken by )(xf asxapproaches ax .
Figure 1
Notice that x approaches ax from two directions. If x approaches )( ax from
right side, then the limit of )(xf asxapproaches ax from right side is written as
)(lim xfax
If )(xf approaches a numberLasxapproaches ax from right side, then we writeas
Lxfax
)(lim
On the other hand, if x approaches ax from left side, then the limit of )(xf as x
approaches ax from left side is written as
)(lim xfax
If )(xf approaches a numberMas xapproaches ax
from left side, then we writeas
Mxfax
)(lim
Consequently, if )(lim xfax
)(lim xfax
, then we say that the limit of )(xf as x
approaches ax exist and denoted by lim ( )x a
f x
. That is, if ML , then
Lxfax
)(lim . If not, we say that )(lim xfax
does not exist.
x
f(x)
from right side
from left side
a
.
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1.2 One Sided Limits
Definition 1.1: Left hand Limit
We write
Mxfax
)(lim
and say the left-hand limit of ( )f x as x approaches a (or the limit of ( )f x as x
approaches afrom the left) is equal of Mif we can make the vales of ( )f x arbitrarily
close toMby takingxto be sufficiently close to aandxless than a(x a ).
Definition 1.2: Righthand Limit
We write
Lxfax
)(lim
Provided we can make ( )f x as close toLas we want for allxsufficiently close to a
and x a without actually lettingxbe a.
Example 1.1:
Evaluate 2lim2 xx
Solution
Note that we are only interested on the behaviour of 2)( xxf asx approaches 2 .
For intuitive understanding, we consider values close to number 2 as given below.
Figure 2
x 2)( xxf
2.2 2.4
1.2 1.4
01.2 01.4
001.2 001.4
0001.2 0001.4
2 4 9999.1 9999.3
999.1 999.3
99.1 99.3
8.1 8.3
From right side
From left side
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From the right side, 2)( xxf approaches value 4. Hence, 42lim2
x
x
.
From the left side, 2)( xxf also approaches value 4. Hence, 42lim2
x
x
.
In conclusion, the limit of 2)( xxf as approaches 2x exist because)(lim
2
xfx
)(lim2
xfx
because and given by 42lim2
xx
.
This is the same value when we substitute 2x directly into the function
2)( xxf . However, this is only true if the function ( )f x is continuous at x a .
In this example, ( ) 2f x x is continuous at 2x as shown of the plot of
( ) 2f x x in Figure 2.
Example 1.2:
Evaluate1
1lim
3
1
x
x
x
Solution
We try to further enhance the understanding intuitively by solving this problem. Note
that 1
1
)(
3
x
x
xf is undefined at 1
x . However, 1
1
lim
3
1
x
x
x can still be found
because our interest is on the neighbourhood ofx only, not on the point 1x . Hence,
we obtained the following table.
Figure 3
x 1
1)(
3
x
xxf
2.1 6400.3
1.1 3100.3
01.1 0300.3 001.1 0030.3
0001.1 0003.3
1 Undefined
9999.0 9997.2
999.0 9880.2
99.0 9701.2
9.0 701.2
8.0 44.2
From left side
From right side
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From the right side,1
1)(
3
x
xxf approaches value 3. Hence, 3
1
1lim
3
1
x
x
x
.
From the left side,
1
1)(
3
x
xxf also approaches value 3. Hence, 3
1
1lim
3
1
x
x
x
.
In conclusion, the limit of1
1)(
3
x
xxf as x approaches 1x exist because
)(lim1
xfx
)(lim1
xfx
because and given by 31
1lim
3
1
x
x
x
. The behaviour of
1
1)(
3
x
xxf when x approaches 1x can be observed easier from Figure 3. It is
clear thatxapproaches 1x , 1
1
)(
3
x
x
xf approaches 3.
1.3 Limits Properties
Table 1: List of basic properties of limits
Limit Properties Example
(a) limx c
a a , where aand cis any
real number
5lim 7 7x , 3lim 2 2x , 8lim 0 0x
(b) lim n nx c
x c
, n is positive integer3 3
2lim 2 8x
x
(c) lim ( ) ( ) lim ( ) lim ( )x c x c x c
f x g x f x g x
3 31 1 1
lim lim limx x x
x x x x
3
1 1 2
3 31 1 1
lim lim limx x x
x x x x
1 1 0
(d) lim ( ) ( ) lim ( ) lim ( )x c x c x c
f x g x f x g x
4 4
2 2 2lim 1 lim lim 1x x x
x x x x
42 2 1 16
(e)lim ( )( )
lim( ) lim ( )
x c
x c
x c
f xf x
g x g x
,
if lim ( ) 0x c
g x
22
0
0
0
limlim
5 lim( 5)
x
x
x
xx
x x
0
05
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(f) lim ( ) lim ( )x c x c
af x a f x
, where a is
any constant
3 2 3 2
2 2
3 2
2 2 2
lim 3 10 3 lim 10
3 lim lim lim 10
3 8 4 10 42
x x
x x x
x x x x
x x
(g) lim ( ) lim ( )n nx c x c
f x f x
(where n is any integer and lim ( ) 0x c
f x
)
OR
1
lim ( ) lim ( )n nx c x c
f x f x
1
lim ( ) n
x cf x
2 233
3 3lim 4 lim 4x x
x x x x
3 39 3 4 8 2 OR
1
32 23
3 3lim 4 lim 4x x
x x x x
1
38 2
(h) lim ( ) lim ( )
nn
x c x cf x f x
, where nisreal number
2
2 2lim 1 lim 1 1x x
x x x
2 2
lim 1 lim 1
3 3 9
x xx x
Example 1.3:
Evaluate the following limits:
(a) xxx
3lim 2
2
(b) 13lim0
xx
(c) 2
24
lim7
x
x
x (d) )43(2lim
25
2 xxx
Solution
(a) xxx
3lim 2
2
2
2
lim xx
+ xx
3lim2
property (c)
2
2
lim xx
2
3 limx
x
property (f)
)2(3)2( 2
2
(b) 13lim0
xx
)13(lim0
xx
property (g)
1limlim300
xx
x properties (c) and (f)
1)0(3 property (a)
1
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(c)2
24lim
7
x
x
x )2(lim
)24(lim
7
7
x
x
x
x property (e)
7 7
7 7
4 lim lim 2
lim lim 2
4(7) 2
7 2
3.33
x x
x x
x
x
property (c) and (f)
(d) )43(2lim 25
2
xxx
)43(lim2lim 22
5
2
xxxx
property (d)
4limlim3lim22
2
2
5
2
xxx
xx property (c) and (f)
512
)4)2(3()2(2 25
Property (f) states that)(lim
)(lim
)(
)(lim
xg
xf
xg
xf
ax
ax
ax
only if 0)(lim
xg
ax
. What happen
when 0)(lim
xgax
. One possibility is that the limit does not exist as illustrated in the
following examples:
2
1lim
2 xx and
x
x
x
4lim
0
Both limits above do not exist. However, the other possibility is that the limit can still
exist because)(
)(
xg
xfis only undefined at ax but both one-sided limits from the left
and right sides and equal (refer Example 2). This can be obtained by simplifying
)(
)(
xg
xffirst as follow:
(a)Factorized )(xf or )(xg and then simplify)(
)(
xg
xf
For example: if 4)( 2 xxf , then )2)(2(42 xxx or
1)( 3 xxf , then )1)(1(1 23 xxxx
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(b)Multiply)(
)(
xg
xfwith the conjugate of )(xf or )(xg and then simplify
For example: if 24)( xxf , then 24)( xxf is the conjugate of
)(xf .
This is illustrated in the following examples.
Example 1.4:
Evaluate
(a)
4
2lim
22 x
x
x
(b)
x
x
x
24lim
0
Solution
(a)
)(
)(lim
2 xg
xf
x
4
2lim
22 x
x
x
Note that 4)( 2 xxg and this can be factorized as )2)(2(42 xxx .
Hence,
4
2
lim 22 x
x
x
)2)(2(
2
lim2 xx
x
x
2
1 1lim
2 4x x
(b)
)(
)(lim
0 xg
xf
x
x
x
x
24lim
0
Note that 24)( xxf and its conjugate 24x . Now, multiplying
)(
)(
xg
xfwith
24
24
x
xdoes not change the function as 1
24
24
x
x. Hence,
x
x
x
24lim
0
24
2424lim
0 x
x
x
x
x
24
44lim
0 xx
x
x
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24lim
0 xx
x
x
4
1
24
1
lim0
xx
1.4 Limits Involve Infinity
In the previous part of this section, the limit of a function )(xf takes a real
number L as xapproaches ax . There is a possibility that as x approaches ax ,
)(xf becomes unbounded, that is )(xf becomes larger and larger towards infinity. If
the function )(xf increases without limit whenxapproaches ax , then we write
)(lim xfax
The above limit takes this form if both the left and right sides of the limit approaches
. That is,
)(lim xfax
and
)(lim xfax
. It is clearly illustrated by Figure 4.
Figure 4
Likewise, the above limit takes this form if lim ( )x a
f x
and lim ( )x a
f x
and
shown graphically in Figure 5. Note that when we write lim ( )x a
f x
or,
lim ( )x a
f x
, that does not mean that the limit exist. This is because or is
not real number. Instead, it only points out that ( )f x will take a very value when x
approaches ax .
( )f x
x
a
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Figure 5
Example 1.5:
Evaluate 1
1
lim 1x x
Solution
Consider the following table.
Figure 6
It can be seen that as xapproaches 1 from right side,1
( )1
f xx
becomes larger
and larger approaching infinity. On the other hand, asxapproaches 1 from left side,
1( )
1f x
x
becomes smaller and smaller approaching negative infinity as shown in
Figure 6. Hence, we can write both limits as 1
1
lim 1x x and 1
1
lim 1x x
x 1
( )1
f xx
0.8 5
0.9 10 0.99 100 0.999 1000 0.9999 10000
1 Undefined1.0001 10000 1.001 1000 1.01 100 1.1 10 1.2 5
x
( )f x
a
From left side
From right side
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respectively. As both one-sided limits have different sign for infinity,1
1lim
1x x
cannot be written as equal to or . In cases above, we only consider the limits
asxapproaches a real number only. It is possible to look at the limits asxapproachesa very large value, usually denoted by symbols or . Those limits are written
as
lim ( )x
f x
and lim ( )x
f x
.
If the function ( )f x approachesL whenxincrease without limit to positive infinity,
then we can write it as
lim ( )x
f x L
If the function ( )f x approachesM whenxdecrease without limit to negative infinity,
then we can write it as
lim ( )x
f x M
These limits can be understood intuitively as well and is illustrated in the next
example.
Example 1.6:
Evaluate1
limx x
Solution
Consider the following table to determine1
limx x
.
x 1
( )f xx
... ...
100000 0.00001
10000 0.0001
1000 0.001
100 0.01
10 0.1
... ...
( )f x
x
( ) 0f x
x
Figure 7
To positive infinity
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Note that as xgoes to infinity,1
( )f xx
gets closer and closer to 0 . Hence, we write
the limits as1
lim 0x x
and illustrated in Figure 7.
Example 1.7:
Evaluate1
limx x
Solution
Consider the following table to determine1
limx x
.
Note that as xgoes to infinity,1
( )f xx
gets closer and closer to 0 . Hence, we write
the limits as1
lim 0x x
and illustrated in Figure 8. We can also find the limits
involving infinity using the basic properties of limits as given Table 1.
Example 1.8:
Evaluate the following limits:
(a) 2lim ( 1)x
x x
(b) 2lim 3( 1)x
x
(c)1
lim4 3x x
Solution
(a) 2lim ( 1)x
x x
2lim lim ( 1)x x
x x
property (a)
x 1
( )f xx
... ...
10 0.1 100 0.01 1000 0.001 10000 0.0001 100000 0.00001
... ...
( )f x
x
( ) 0f x
x
Figure 8
To negative
infinity
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2lim lim lim 1x x x
x x
property (c)
( 1)
( )
(b) 2lim 3( 1)x
x
23 lim ( 1)x
x
property (f)
3
(c) 1lim4 3x x
lim 1
lim 4 3x
xx
property (e)
lim 1
4 lim lim 3
x
x xx
properties (c) and (f)
1
(4 ) 3
1
0
From example 1.8, we use standard results involving infinity such that (4 ) or
1 or ( ) or4
or
10
. However, what happen if the
answer for( )
lim( )x
f x
g xis
? The alternative approach is to change the form of
( )
( )
f x
g x
first by multiplying( )
( )
f x
g xwith
1
1
n
n
x
x
where nis the highest power ofxin ( )g x .
This is illustrated in the next examples.
Example 1.9:
Evaluating 22 3
lim1x
x
x
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Solution
Direct substitution of in the function gives2
2 3lim
1x
x
x
. Hence, let us consider
( )lim( )x
f xg x
22 3lim 1xx
x
.
Now, the term with the highest power in 2( ) 1g x x is 2x . Then, we multiply
( )
( )
f x
g x 22 3
1
x
x
with
2
2
1
1x
x
, so that
( )( )
f xg x 2
2 31
xx
2 2 2 2
2
2 22 2
1 2 3 2 3
1 111
x
x x x x xx
x xx x
Hence,2
2 3lim
1x
x
x
2
2
2 3
0 0lim 0
1 1 01
x
x x
x
Example 1.10:
Evaluate the following limits:
(a)2
2
3 1lim
1x
x
x
(b)
35 1lim
1x
x
x
(c)
2
3 1lim
1x
x
x
Solution
(a)2
2
3 1lim
1x
x
x
2
2
3 1lim
1x
x
x
2
2
1
1
x
x
2
2
2
2
2
2
3 1
lim1
13
3 0lim 3
1 1 01
x
x
x
x
x
x
x
x
METHOD NOTE
2x is the with the highest
power ofxin the denominator.
Hence, we multiply
( )
( )
f x
g xwith
2
2
1
1
x
x
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(b)35 1
lim1x
x
x
35 1lim
1x
x
x
1
1
x
x
35 1
lim1x
x
xx
x
2 15
0lim
1 1 01
x
xx
x
(c)2
3 1lim
1x
x
x
23 1
lim1x
x
x
2
2
1
1
x
x
2
2 2
3 1
lim1x
x
x x
x
x x
2
13
3 0lim 3
1 1 01
x
x
x
Exercise 1.1:
Evaluate the following limits:
(a)3 1
lim1x
x
x
Answer: 0
(b)1
lim1x
x
x
Answer: 0
METHOD NOTE
x is the with the highestpower ofxin the denominator.
Hence, we multiply
( )
( )
f x
g xwith
1
1
x
x
METHOD NOTE
2x is the with the highest
power ofxin the denominator.
Hence, we multiply
( )( )
f xg x
with2
2
1
1
x
x
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TUTORIAL 1
Limits
1.
Evaluate the following limits
(a) 1023lim 231
xxxx
(b) 41lim 222
xxx
(c)1
23lim
2
2
3
x
xx
x (d) 3 2
3215lim
xx
x
(e) 9
710
lim 2
3
5
x
xx
x (f) 6
6
lim6
x
x
x
2.
Find the following limits
(a) 73lim2
xx
(b) 37lim
xcx
(c) )1)(2(lim1
xxx
(d))1(
)2(lim
0
x
x
x
(e) 2lim5
xx
(f) 31
5)74(lim
x
x
(g)9
1lim
23 xx (h)
9
3lim
23
x
x
x
(i)592
5lim
25
xx
x
x (j)
27
9lim
3
2
3
x
x
x
(k) 23lim 23
xx
(l)
x
x
x
5
lim3
3. Find the limits in question (a)(d).
(a) )(lim2
xfx
, )(lim2
xfx
and )(lim2
xfx
where
23
227)(
xx
xxxf
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(b) )(lim2
xfx
, )(lim2
xfx
and )(lim2
xfx
where
0
0)(
xx
xexf
x
(c ) )(lim4
xfx
, )(lim0
xfx
and )(lim3
xfx
where
43
402
01
)(
x
x
x
xf
4. Evaluate the limits, if it exists (Hint: Use One-Sided Limits)
(a) 3lim3
xx
(b)a
x
ax
2lim
(c)
2
2lim
2
x
x
x (d) 3lim
4x
(e)
16
42lim
16
x
xx
x (f)
4
8lim
2
3
2
x
x
x (g) 8lim
x
5.
Find the limits
(a) 4
2
2 24lim
x
xx
(b)12
lim
2
1 xx
x
(c)
2
4 1lim3 1x
x xx
(d)1
42lim
2
2
x
x
xx (e)
2
6lim
2
2
x
xx
x (f)
25
5lim
25
x
x
x
(g)2
11lim
2
x
x
x (h)
2
8lim
3
2
x
x
x (i)
x
x
x
2
16lim
2
4
(j)
x
x
x
52
1lim
4 (k)
12
124lim
3
23
xx
xxx
x (l)
4
2lim
22
x
x
x
(m)
xxx
11lim
2 (n)
2
1lim
xx (o) 13lim 2
x
x
(p)12
lim2
x
x
x (q)
22
3lim
x
x
x (r)
32
2lim
2
x
xx
x
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1.5 Continuity
Continuity, in layman word, means smooth transition from something to
another. Continuity in function means the motion of the function from one point to
another is unbroken. Intuitively, we can imagine drawing such continuous function
using a pen without having to lift the pen smoothly. For example, consider plots in
Figure 9.
Figure 9 (a) Figure 9 (b)
Figure 9 (c) Figure 9 (d)
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Consider Figure 9 (a) and 9 (b). If we move a pen along the curve from left to right,
we do not have to lift the pen at all. That means, the function is continuous at x a .
However, for Figures 9 (c) and 9 (d), we have to lift the pen at x a because thefunction is undefined at x a . Thus, these functions are not continuous.
Definition 1.3: Continuous
A function ( )f x is said to be continuous at x a if it satisfies the following three
conditions.
(a) ( )f a is defined (exist)
(b) lim ( )x a
f x
exists if L M where
lim ( )x a
f x L
and lim ( )x a
f x M
(c) lim ( ) ( )x a
f x f a
Example 1.11:
Determine whether23 5
( )2 1
xf x
x
is continuous at 3x .
Solution
We inspect the three conditions:
(a) Is (3)f defined?
23(3) 5 32
(3) 4.5714 4.572(3) 1 7f
. Hence, (3)f is defined and the first
condition is satisfied.
(b)Does3
lim ( )x
f x
exist?
Right-hand limit: take 3.001x ,2
3
3(3.001) 5lim 4.57269 4.57
2(3.001) 1x
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Left-hand limit: take 2.999x ,2
3
3(2.999) 5lim 4.5701 4.57
2(2.999) 1x
Since3 3
lim ( ) lim ( ) 4.57x x
f x f x
. Hence, the limit exists and the second
condition holds
(c)
Does ( ) (3)f x f ?
From (a) and (b), the third condition also hold. Hence, we conclude that
23 5( )
2 1
xf x
x
is continuous at 3x .
Example 1.12:
Determine whether
2 ; 0( )
2 ; 0
x xf x
x x
is continuous at 0x .
Solution
We inspect the three conditions:
1. Is (0)f defined?
2(0) (0) 0f . Hence, (0)f is defined and the first condition is satisfied.
2.
Does0
lim ( )x
f x
exist?
Right-hand limit: take 0.001x , 20
lim(0.001) 0.000001 0x
Left-hand limit: take 0.001x ,0
lim 2( 0.001) 0.002 0x
Since0 0
lim ( ) lim ( ) 0x x
f x f x
. Hence, the limit exists and the second condition
holds.
3. Does ( ) (0)f x f ?
From (a) and (b), the third condition also hold. Hence, we conclude that ( )f x is
continuous at 0x .
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Exercise 1.2:
Determine whether the function given is continuous at the indicated point.
(a)2 ; 1
( )
; 1
x xf x
x x
is continuous at 1x . Answer: not continuous
(b)
4 12
( ) 2
3 2 2
xx
f x x
x x
is continuous at5
3x . Answer: continuous
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TUTORIAL 2
Continuity
1.
Determine whether the function given is continuous at the indicated point.
(a) 52)( 3 xxf ; 2x (b) 3)( 2 xxf ; 4x
(c)3
2)(
x
xxf ; 3x (d)
31
3)(
2
2
xx
xxxf ; 3x
(e)
11
11)(
2
xx
xxxf ; 1x (f)
10
11
12
)( 2
x
xx
xx
xf ; 1,1x
2. Determine whether or not the function is continuous at the indicated point.
(a) 2;15)( 3 xxxxf
(b) 1;51)( 2
xxxg
(c) 2;2,
;2,4)(
3
2
x
xx
xxxf
(d) 2;2,
;2,5)(
3
2
x
xx
xxxf
(e) 2
;2
;25
;2,4
)(
3
2
x
xx
x
xx
xf
(f) 2
;21
;210
;2,5
)(
3
2
x
xx
x
xx
xf
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3. A functionfis defined as follows:
1 1
( ) 1 1 1
1 1
x
f x kx x
x x
(a)Given thefis continuous at 1x . Find the value of constant k.
(b)Determine whetherfis continuous at 1x .
(c)Sketch the graph of ( )y f x
4. A functionfis defined as follows:
24
2( ) 2
3 2
xx
f x xx
(a)Sketch the graph
(b)Discuss the continuity offat 2x .