chapter 1 - limits.pdf

7/28/2019 Chapter 1 - Limits.pdf

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SFM 1002 Mathematics II

Chapter 1 Limits

1

CHAPTER 1

LIMITS

________________________________________________________________________

At the end of chapter, the students are able to

find the limit of function )(xf , whetherx approaches a real number or (C1) trace the limit properties to solve the limit problems (P3, C3) use the concept of limit to determine continuity of the function (C3)

________________________________________________________________________

1.1 Introduction to Limits

Definition 1.1: Letx be any real number variable, and let c be a fixed real number. To say thatxapproaches c

means thatx can take on any value which is arbitrarily close to c.

Notation: x approaches c is written as cx . If cx , then the distance between x and c

differs by arbitrarily small amount which can be denoted by a positive number such that


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Chapter 1 Limits

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Example 1.2(a): Let () = 29

3and 3. As we known, () is defined for all except

= 3. To find the limit of() when approaches 3, we choose any number close to 3.

2 2.5 2.9 2.99 2.999

() 5 5.5 5.9 5.99 5.999

or

4 3.5 3.1 3.01 3.001 3.0001

() 7 6.5 6.1 6.01 6.001 6.0001

Figure 2 : Table ofx and ()

From the table above we can observe that asx approaches 3, () will approach 6. If we choose

x close enough to 3, thenf will be nearer to 6. We conclude that 6 is the limit off asx

approaches 3.

The above result can be written as follows,

lim3

2 9

3= 6.

Example 1.2(b): Let5

125)(

3

=

x

xxf and 5. Find the limit of )(xf when approaches 5.


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Chapter 1 Limits

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Definition 1.3: The limit of() when approaches from right side can be written as

lim +

().

Definition 1.4: The limit of() when approaches from left side can be written as

lim ().

Proposition 1.1: If Lxfcx

=+

)(lim and Lxfcx

=

)(lim , then Lxfcx

=

)(lim .

Example 1.3(a): Find 12lim3

+

xx

.

Solution:

Step 1: 712lim3

=++

xx

;

Step 2: 712lim3

=+

xx

.

Conclusion: Because of 12lim12lim33

+=++

xxxx

, thus 712lim3

=+

xx

.

Example 1.3(b): Find 1lim2

5+

x

x.

Proposition 1.2: If )(lim)(lim xfxfcxcx+

, then )(lim xfcx

does not exist.

Method:

To find the limit of the function )(xf , we need to the follow:

Step 1: Find the value of )(lim xfcx+

Step 2: Find the value of )(lim xfcx

Step 3: Make a conclusion

If )(lim)(lim xfxfcxcx +

= , thus )(lim)(lim)(lim xfxfxfcxcxcx+

==

If )(lim)(lim xfxfcxcx +

, thus )(lim xfcx

does not exist.

Note: The function )(xf and the number c will be given.


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Chapter 1 Limits

4

Example 1.4(a): Find )(lim2

xfx

, where


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Chapter 1 Limits

5

Example 1.5(a): Let 2=n , 5=k , 3=c ,3

9)(

2

=

x

xxf , and 12)( += xxg . Verify the limit

properties.

Solution:

Step 1: 639lim)(lim

2

33 == xxxf

xxand 712lim)(lim

33 =+= xxg xx ;

Step 2:

1. 55limlim3

== xcx

k ;

2. 3065)(lim5)(lim)(lim3

====

xfxfkxkfxcxcx

;

3. [ ] 1376)(lim)(lim)(lim)(lim)()(lim33

=+=+=+=+

xgxfxgxfxgxfxxcxcxcx

;

4.

[ ] 176)(lim)(lim)(lim)(lim)()(lim 33====

xgxfxgxfxgxf xxcxcxcx ;

5. [ ] 4276)(lim)(lim)(lim)(lim)()(lim33

====

xgxfxgxfxgxfxxcxcxcx

;

6.7

6

)(lim

)(lim

)(lim

)(lim

)(

)(lim

3

3 ===

xg

xf

xg

xf

xg

xf

x

x

cx

cx

cx;

7. [ ] 366)(lim)(lim)(lim 223

====

xfxfxfx

n

cx

n

cx;

8. 223

6)(lim)(lim)(lim ===

xfxfxfx

ncx

n

cx.

Example 1.5(b): Let 3=n , 7=k , 5=c ,5

125)(

3

=

x

xxf , and 1)( 2 += xxg . Verify the limit

properties.


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Chapter 1 Limits

6

1.2 Limit involve Infinity

Definition 1.5: If the function )(xf increases without limit when x approaches c , then we

write +=

)(lim xfcx

.

)(xf

+ +

x

c

Figure 3: Limit )(xf approach +

Definition 1.6: If the function )(xf decreases without limit when x approaches c , then we

write =

)(lim xfcx

.

)(xf

c x

Figure 4: Limit )(xf approach

Notation: +=

)(lim xfcx

or =

)(lim xfcx

does not mean that limit does not exist. This is

because + and is not real number.

Example 1.6(a): Find3

1lim

3 xx.


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Chapter 1 Limits

7

Solution:

Step 1: +=+ 3

1lim

3 xx;

Step 2: +=

3

1lim

3

xx

.

Conclusion: Because of3

1lim

3

1lim

33 =

+ xx xx, thus +=

3

1lim

3 xx.

Example 1.6(b): Find1

1lim

1 xx.

Definition 1.7: If the limit of the function )(xf as x approaches a very large value, like + and then we write )(lim xf

x +and )(lim xf

x .

Proposition 1.3: 01

lim =+ nx x

for any number of n .

Method:

To find the limit of the function )(xf as x approaches + or , we need tothe follow:

Step 1: Suppose that)(

)()(

xq

xpxf = , then determine the highest order of

x for functions )(xp and )(xq , let say is n ;

Step 2: Both functions )(xp and )(xq divided by nx ;

Step 3: Using the limit properties to solve the limit of function )(xf


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Chapter 1 Limits

8

Example 1.7(a): Find52

34lim

+

++ x

x

x

Solution :

22

4

0.52

0.34

(3))&(2)1.1Theorem(1

lim52lim

1lim34lim

(6))1.1Theorem(5

2lim

34lim

)bydivide(52

34

lim52

34lim

==+

+=

+

+

=

+

+

=

+

+=

+

+

++

++

+

+

++

x

x

x

x

x

x

x

x

x

xx

xx

x

x

xx

Example 1.7(b): Find14

52lim

3

2

+ x

xx

x

Example 1.7(c): Find52

43lim

2

++ x

x

x


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Chapter 1 Limits

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1.3 Continuity

Definition 1.8: A function )(xf is said to be continuous at cx = if )(xf satisfy the following

three conditions.

(a) )(cf is defined (exist);(b)exist )(lim xf

cx, that mean )(lim)(lim xfxf

cxcx+

= ;

(c) )()(lim cfxfcx

=

.

Example 1.8(a): Determine whether12

34)(

2

+

+=

x

xxf is continuous at 2=x .

Solution:

Step 1:5

19

1)2(2

3)2(4)2(

2

=+

+=f

Thus, )(cf is defined.

Step 2:5

19

12

34lim)(lim

2

2=

+

+=

++ x

xxf

xcx

5

19

12

34lim)(lim

2

2=

+

+=

x

xxf

xcx

Thus, exist )(lim xfcx

, because )(lim)(lim xfxfcxcx+

=

Step 3: )2()(lim2

fxfx

=

Conclusion: )(cf is continuous at 2=x .

Method:

To determine the function )(xf is continuous at cx = , we need to the follow:

Step 1: Find )(cf , the value of )(cf must be a real number;

Step 2: Find the values of )(lim xfcx+

and )(lim xfcx

, if they are equal, then

exist )(lim xfcx

;

Step 3: Check )(cf is equal to )(lim xfcx

or not.

Conclusion: If all conditions are fulfill, then function )(xf is continuous at

cx = ; if one of them cannot fulfill, then function )(xf is not

continuous at cx = .


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Chapter 1 Limits

10

Example 1.8(b): Determine whether3

13)(

2

++=

x

xxxf is continuous at 3=x .

Exercise 1:

1. Evaluate the following limits:

a) )11392(lim235

2+

xxx

xb)

695

73lim

2

3

1 ++

++ zz

zz

x

c)3 2

2 23lim xxx d) )73(lim2

2 + xxx

e)4

2lim

4

x

x

xf)

2

6lim

2

2

+ x

xx

x

g) )45(lim33

0+

tt

xh) )72)(5(lim

3+

xx

x

i)1

3lim

2

1 +

+ z

zz

xj)

753

103lim

2

2

3 +

+ xx

xx

x

k)14x19x4x

6x11x6xlim

23

23

1x ++

+

l)

2x

10x7xlim

2

2x +

++

m)1x

2xxlim

2

2

1x

+

n)21x4x

51x14xlim

2

2

3x

o)6uu

x2u2uxulim

2

2

2u

+

p)

3x2x

uxuxxlim

2

2

1x +

+

q)22

22

x x

4x6x2lim

+

r)

4w4w

)6ww)(2w(lim

2

2

2w ++

+


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Chapter 1 Limits

11

2. Find the following limits:

a)5x

xlimx

b))x3)(5x(

xlim

2

x

c)3

2

x x5

xlim

d)

15x8x

xlim

2

2

x +

e)2

2

t t7

tlim

f)

23

3

x x100x2

xlim

g)5t

tlimt

h)45

5

5lim

3. Determine the function )(xf is continuous or not

a) 1)( 2 ++= xxxf at 3=x b)1

4)(

2

+

=

x

xxf at 2=x

c)1

22)(

3

+=

x

xxxf at 1=x d)

1

22)(

3

+

++=

x

xxxf at 1=x

e)4

16)(

2

=

x

xxf at 4=x f)

1

43)(

2

+

=

x

xxxf at 4=x

g)

=

+=

0for,0

0for,2)(

x

xxxf at 0=x i)

chapter 1 - limits.pdf

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