calculus limits notes(2014)

VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2 i

MA 125- Calculus

Table of Contents Page

1.Definition of Limits

Left hand and right hand limit

Calculating the limit at a point.

2..Exercise # 1

4 Limit Theorems

5 Solution to Exercise # 1

6 Exercise # 2

7.Finding Limits of Functions

In Class Assessment

9 Exercise # 3-4

10 Solution-Exercise # 4

11 Exercise #5

12Continuity

14Worksheet # 1

15 Worksheet # 2

VTDI- SEPS-Humanities-Calculus- Limits Handout- June 2014.V1.2 1

Module 1.2- Limits Definition The limit of a function f(x), is the value that f(x) approaches, as x tends towards a. This is

written as ax

lim f(x) = L. This means that as x gets closer to a, from either side (but not

equal to a) the value of f(x) gets closer to L.

Suppose that our function f(x) is x2

+ 2. If one needs to find the value of the limit of f(x) at

x equal 2. We need to find out the value of f(x) as x approaches 2 from above and below.

This is easily done by observing a table of values.

x 1.70 1.75 1.80 1.85 1.90 1.95 1.99

f(x) 4.89 5.0625 5.24 5.4225 5.61 5.8025 5.9601

Notice for this table, 2 is approached from the left hand side. If 2, is approached from the

right hand side the following table illustrates the result.

x 2.30 2.25 2.20 2.15 2.10 2.05 2.01

y 7.29 7.0625 6.84 6.6225 6.41 6.2025 6.0401

From the tables above, one should realize that as x approaches 2 the value of x2 +2 tends

towards 6.

Left hand limit & Right hand limit.

The left hand limit is the value that f(x) approaches as one tends to where x = a from the

left hand side. From the example above, we approached 2 from the left, this may be

written as )2(lim 22

xx

= 6. If two was approached from the right hand side it would be

expressed as, 6)2(lim 22

xx

.

Please note that the left hand and right hand limit are equal. The limit only exists if the

left hand and right hand limit are equal.

Lets investigate for f(x) = 2

232 2

x

xx. By using the table of values below one may be

able to investigate for 2

limx 2

232 2

x

xx.

x 1.75 1.80 1.85 1.90 1.95 1.99

y 4.5 4.6 4.7 4.8 4.9 4.98


x 2.25 2.20 2.15 2.10 2.01

y 5.52 5.4 5.3 5.2 5.02

By examining the tables it can be seen that it is true to say that as x tends to 2 the value of

f(x) approaches 5. Hence, the limit as x tends to 2 is 5. This may be written as

2limx

2

232 2

x

xx= 5.

This method of calculating limits is tedious; hence the alternative is to calculate limit the

limit at the respective points. The method of calculation used, is dependent on the type of

function and also the value that the limit tends to.

In this section we will discuss the most suitable method for calculating the limits of some

functions.

Find the limits of the following functions.

1. )23(lim

2

x

x. The limit of this function may be determined by direct substitution.

)23(lim2

xx

= 3(2)-2 =6-2 = 4

2. )42(lim 22

xxx

By direct substitution, implies 2(2)2 -4(2)= 8 -8 = 0

N.B. Direct substitution is used for functions that are determined by

polynomials.

3. 1

lim2

1

x

xx

x, Notice that if direct substitution is used with this question, the result

will be undefined. This solution is not true. The real solution is determined by

factorization, )1(

)1(lim

1

x

xx

x= x

x 1lim

.

Notice that this rational expression is simplified to a linear function, xx 1lim

=1.

This strategy is employed when the direct substitution method fails especially

when the function is a rational function.

4. 2

3lim

22 xx By direct substitution

2

3lim

22 xx =

22

32

=6

3


This question posed little difficulty since its not a function of the format )(

)(

xg

xf.

In other words it isnt a rational function so direct substitution could be used to evaluate the limit.

5. 3

34lim

2

3

x

xx

x.

For this question direct substitution fails again. The rational function may be

reduced by using factorization to simplify the expression.

3

34lim

2

3

x

xx

x =

)3(

)1)(3(lim

3

x

xx

x= )1(lim

3

x

x= -3 +1= -2

Exercise 1.

Compute the limits of the functions.

1. )61(lim

1x

x

2. x

x

x

5

1lim

2

5

3. x

xx

x

3lim

2

0

4. 1

1lim

2

1

x

x

x

5. 2

42lim

2

2

x

xx

x

6. 3

6lim

2

3

x

xx

x

7. x

x

x

4

16lim

2

4

8. 25

102lim

25

x

x

x

9. 65

6lim

2

2

6

xx

xx

x

10. )11392(lim 2352

xxxx

11. )73(lim 22

xxx


12. )72)(5(lim

)45(lim

3

23

0

xx

tt

x

x

13. 3 22

23lim

xxx

14. 3

9lim

2

3

x

x

x

15. x

x

x

1)1(lim

2

0

16. 9lim 24

xx

Limit Theorems

If the limit of f(x) and g(x) exist, that is if Mxfax

)(lim and .)(lim Nxgax

I. ))()((lim xgxfax

= )(lim xfax

+ )(lim xgax

= M +N

II. )(lim xcfax

=c )(lim xfax

=cM

III. )()(lim xgxfax

= )(lim)(lim xgxfaxax

= MN

IV. )(

)(lim

xg

xf

ax=

)(lim

)(lim

xg

xf

ax

ax

= N

M. This is true only if )(lim xg

ax 0

V. ccax

lim

VI. 2))((lim xfax

= 2))(lim( xfax

= M2

Please note that there exist other theorems of limits that the reader will find useful.

Limits as one Approaches Infinity

Calculate 1

1lim

x

x

x.

If one uses direct substitution, it would result in a large numerator and a large

denominator. It is not mathematically sound to say infinity divided by infinity gives 1.

Hence, the solution is accomplished by dividing each term within the numerator and

the denominator by the highest power of the variable within the denominator.


Hence the result will be,

xx

xxx

x

x 1

1

lim

. This is simplified to,

x

xx 1

1

11

lim

.

Please note that x

1 approaches zero as x tends towards infinity hence the limit of the

function is 1.

Eg. 2. Calculate 2

53lim

2

2

x

x

x

22

2

22

2

2

53

lim

xx

x

xx

x

x

=

xlim

2

2

21

53

x

x

,

By substitution the result will be 3 since dividing by infinity result is zero.

Solution to Exercise 1.

1. 1-6= -5

2.

0

26

55

152

3. 3)3(lim)3(

lim00

x

x

xx

xx

4. 2)1(lim)1(

)1)(1(lim

11

x

x

xx

xx

5. 42lim)2(

)2(2lim

22

x

x

xx

xx

6. 5)2(lim)3(

)2)(3(lim

33

x

x

xx

xx

7. 8)4(lim)4(

)4)(4(lim

)4(

)4)(4(lim

444

x

x

xx

x

xx

xxx

8. 5

1

)5(

2lim

)5)(5(

)5(2lim

55

xxx

x

xx


9.7

6

1lim

)1)(6(

)6(lim

66

x

x

xx

xx

xx

10.64-72 + 12-11= -7

11. 4 +6 -7 = 3

12.-8

13. 28264 33

14. 6)3(lim)3(

)3)(3(lim

33

x

x

xx

xx

15. 22lim2

lim121

lim0

2

0

2

0

x

x

xx

x

xx

xxx

16. 59lim 24

xx

Exercise 2.

Evaluate the following limits.

1. x

limx

x

21

2. 9

62lim

2

x

xx

x

3. x

x

x 21lim

4. 2

1lim

xx

5. 2

12lim

x

x

x

6. 1

121512lim

2

2

t

tt

x

7. 13

68lim

x

x

x

8. 13

23lim

2

x

x

x

9. 34

23lim

2

x

x

x


10. 123

45lim

2

32

xx

xx

x

Finding Limits by Rationalizing

Evaluate the limit of h

h 11 as h tends to zero.

Method - To rationalize the expression, both numerator and the denominator must be

multiplied by the conjugate of the numerator. The conjugate for this example is

h1 +1.

0limh h

h 11 x

11

11

h

h=

)11(

11

hh

h=

)11( hh

h

=0

limh 11

1

h =

2

1

Attempt

7

32lim

7

x

x

x N. B. The conjugate of the numerator is 2x +3

Answer = 6

1

In Class Assessment Calculus

1. Given that )(lim xfax

=-3, )(lim xgax

= 0 & )(lim xhax

= 8.

Find the limits that exist. If the limit does not exist explain why.

a) ax

lim [f(x) +h(x)]

b) ax

lim [f(x)]2

c) ax

lim 3 )(xh


d) )(

)(lim

xh

xf

ax

e) )(

)(lim

xh

xg

ax

f) )(

)(lim

xg

xf

ax

g) )()(

)(2lim

xfxh

xf

ax

2. Evaluate the following limits.

a) 4

limx

(5x2 -2x +3)

b) 2

limt

(t +1)9(t2 -1)

c) 2

limuu4 +3u + 6

d)

372

9lim

2

2

3

tt

t

t

e) h

h

h

8)2(lim

3

0

3. Show that

12

54lim

2

2

x

xx

x = 2

4. Evaluate the following limits

a) 4

53lim

x

x

x

b)

42

5lim

3

3

xx

xx

x

c)

1

2lim

23

2

tt

t

t

5. Sketch a graph that illustrate the following characteristics

)(lim0

xfx

,

)(lim0

xfx

, 1)(lim

xfx

, 1)(lim

xfx


Exercise 3.

Calculus

Answer all questions showing appropriate working

1. Given that ax

lim f(x) = -3, ax

lim g(x) = 0 and ax

lim h(x) = 8

Find the limits of

(a) ax

lim [f(x) + h(x)] (b) ax

lim [f(x)]2

(c) ax

lim )(

)(

xg

xf (d)

axlim

)(

)(

xh

xf

2. Evaluate the limits of the following using the laws of limits where appropriate

(a) 4

limx

(5x2 -2x + 3) (b)

2lim

x 46

122

xx

x

(c) 3

limt 372

9

tt

t (d)

2limx

2

62

x

xx

(f) x

lim 4

53

x

x

Exercise 4.

1. Suppose 3)(lim2

xfx

and 4)(lim2

xgx

. Use the limit rules to find the following

limits.

a. )(5)(lim2

xgxfx

b. )()(2lim2

xgxfx

c. )(lim

)(3lim

2

2

xf

xg

x

x

Find the limits

2. 1

1lim

2

3

x

xx

x


3. 2

6lim

2

2

x

xx

x

4. 4

2lim

4

x

x

x

5. 32

1 1

12lim

x

xx

x

Solution -Exercise 4

6. Suppose 3)(lim2

xfx

and 4)(lim2

xgx

. Use the limit rules to find the following

limits.

a. 23)4(53)(5lim)(lim)(5)(lim222

xgxfxgxfxxx

b. 2432)(lim)(2lim)()(2lim222

xgxfxgxfxxx

c.

43

43

)(lim

)(3lim

2

2

xf

xg

x

x

Find the limits

7. 2

5

4

5

1lim

133

1lim

)1(lim

1

1lim

3

3

3

2

3

2

3

xx

xx

x

xx

xx

x

x

8.

532)3(lim)2(lim

)3)(2(lim

)2(lim

6lim

2

6lim

2

2

2

2

2

2

2

2

x

x

xx

x

xx

x

xx

x

x

x

x

x

x

9.

4

1

22

1

24

1

2

1lim

24

4

24

2

2

2

4lim

2lim

4

2lim

4

22

4

4

4

x

xx

x

xx

x

x

x

x

x

x

x

x

x

x

x

10.

existnotdoes

xxxx

xx

x

xx

x

x

x

xlim

0

1

11

1

1

1lim

111lim

11lim

1

12lim

1

1

1

3

2

1


Exercise 5.

1. Evaluate the following limits

(a) 2

38

lim4

x

xx

(b) 4

16lim

2

4

x

x

x

(c) 2

65lim

2

2

x

xx

x

(d) 43

32lim

2

2

1

xx

xx

x

(e) 3

9lim

9

x

x

x

2. Determine the values of x for which the following functions are

discontinuous.

(a) f(x) = 2

232

x

xx

(b) f(x) = 86

1032

2

xx

xx

3. Given that 3

limx

{f(x) + 3x} =1,evaluate )}(9{lim3

xfx

4. Given that 2

limx

{4f(x)} = 5,evaluate 2

limx

{f(x) + 2x}

5. Find the limits of the following functions

a) 10

43lim

x

x

x

b) 12

63lim

2

x

xx

x


Continuity

A function is continuous at a point where x = c if the following conditions are satisfied.

f(c) is defined

cx

lim f (x) exist

cx

lim f (x) = f(c)

If the function is not continuous it is said to be discontinuous. All functions

which are defined by polynomials are continuous inclusive of any exponential

functions and the log function y = logax for all x> 0.

N.B. When a function is defined by a given formula it is usually continuous for all

values within the domain. If a function has a change in the defining formula then

there exists some value of x, within the domain at which the function is

discontinuous.

Points of Discontinuity

A rational functions point of discontinuity may be easily identified since

this is occurs whenever the denominator of the function is zero. The function is

therefore undefined at this point. A function may also have more than one point of

discontinuity. Hence, one has to explore all the possibilities for the function to be

discontinuous.

In this section we examine two types of rational functions and a piecewise

function in order to determine their points of discontinuity.

Example 1. f(x) = 2

5

x

x and g(x) =

)2(

5

xx

x.

Notice that f(x) is discontinuous at x = 2 while g(x) is discontinuous at x = 0 and

x = 2. For these values of x within the domain the function is undefined.

Example 2. Given the function f(x) = 1 if x < 2

x +3 if 2 x 4 7 if x > 4

This is a piecewise function because it has three different expressions by which it

is defined and each expression uses a specified domain. To check whether the


function is discontinuous, a point within the domain has to be identified where

this occurs. A sketch of this graph will be useful to identify the point where

discontinuity occurs. Make an attempt to draw the graph.

From the sketch, the graph should indicate that f(x) is discontinuous at x = 2. Also

by observations and calculations when x < 2, f(x) =1, but at x = 2, f(x) = 2 +3 =5.

Since there is a jump from f(x) =1 to f(x) = 5, this is a jump discontinuity, f(x) is

discontinuous at x = 2.

It is easy to identify points of discontinuity by observing a graph.

1. If there is point on the graph where the limit does not exist the graph is

discontinuous.(Remember that the limit only exist if the left hand limit

and the right hand limit are equal)

2. If the limit exist at x = c, but it is not equal to f(c) the graph is discontinuous.

3. If f(c) does not exist also the graph is discontinuous.

The following exercise should be useful to strengthen your understanding of

continuity.


a. Hole: f(c) is not defined existsxfcx

)(lim

.

b. Hole: f(c) is defined existsxfcx

)(lim

and is not equal to f(c).

c. Jump: )(lim1

xfcx

is not the same as )(lim xfcx

.

d. Pole : f(c) is defined at x = c;

)(lim1

xfcx

e. Pole: f(c) is defined at x = c ;

)(lim1

xfcx

and

)(lim xfcx

Worksheet #1

Find the limits of the following:

1. xx

xx

x 2

124lim

2

2

2

2.

h

h

h

1832lim

2

0

3. t

tt

t

4

43lim

4

4. x

x

x

33lim

0

5. Evaluate

33

24

312

lim3

xx

xx

xfx

Sketch a graph to represent xfx 3lim

6. Using the graph, compute each of the following.


(i) 4f (ii) 4

lim ( )x

f x

(iii) 1

lim ( )x

f x

7. (a) List the properties of a function which is continuous at a.

(b) Show that f(x) is continuous at x = -1

State the values of x for which the function is discontinuous.

(i) f(x) = 103

22

xx

x (ii) g(x) =

16

42

x

x

8. Determine where the function below is not continuous 152

1042

tt

tth

Activity Sheet

Calculus worksheet #2 - Continuity

a. Is 4

16)(

2

x

xxf continuous at x = 4. Give reason(s) for your answer.

b. Find the value of the constant A that makes the given function f(x) continuous for

all x.

13

11

1

)(2

2

xifxAx

xifx

x

xf

c. Determine if f(x) is continuous at x = -2, where


263

24)(

xifx

xifxxf

d. Determine if f(x) is continuous at x = 2, where

2

2

11

5

23

)(4 xif

xif

x

xifx

xf

END OF HANDOUT

calculus limits notes(2014)

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