chapter 1 limits

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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Chapter 1 Limits

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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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Chapter 1 Limits

Page 4

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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Chapter 1 Limits

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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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Chapter 1 Limits

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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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Chapter 1 Limits

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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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Chapter 1 Limits

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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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Chapter 1 Limits

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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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Chapter 1 Limits

Page 10

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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Chapter 1 Limits

Page 11

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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Chapter 1 Limits

Page 12

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:


(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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Chapter 1 Limits

Page 13

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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Chapter 1 Limits

Page 14

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:


(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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Chapter 1 Limits

Page 15

(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:


(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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Chapter 1 Limits

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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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Chapter 1 Limits

Page 17

(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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Chapter 1 Limits

Page 18

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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Chapter 1 Limits

Page 19

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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Chapter 1 Limits

Page 20

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


Example 1.12:

Determine whether

2 ; 0( )

2 ; 0

x xf x

x x


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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Chapter 1 Limits

Page 21

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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Chapter 1 Limits

Page 22

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


23/23

Chapter 1 Limits

Page 23

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 .

chapter 1 limits

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