Algebra of LimitsAssume that both of the following limits exist and c and is a real number:

Then:

0)(lim

),(lim/)(lim)(/)(lim..4

)(lim)(lim)()(lim..3

)(lim)(lim)()(lim..2

)(lim)(lim..1

xgthatprovided

xgxfxgxf

xgxfxgxf

xgxfxgxf

xfcxcf

ax

axaxax

axaxax

axaxax

axax

Calculating LimitsFinding the limit of a function f a point x = a.

Distinguishing the following cases:1. The case when f is continuous a x = a.2. The case 0/0.3. The case ∞/ ∞4. The case of an infinite limit5. The case c/∞, where c is a real number.6. The case, when it is possible to use the

squeeze theorem.

1. The case when f is continuous at x = a

If f is continues at x=a, then:

Notice:1. Polynomial functions and the cubic root function ( & all functions of its two families) are everywhere continuous.2. Rational, trigonometric and root functions are continuous at every point of their domains.3. If f and g are continuous a x=a, then so are cf, f+g, f-g, fg and f/g (provided that he limit of f at x=a is not zero)

Examples for the case when f is continuous at x = a

04

0

22

4)2()2()(lim

2.},2{

.2

4)(

2

4lim

)1(

2

2

2

2

2

fxf

xatcontisitsoandRon

contisx

xxffunctionrationalThe

x

x

Example

x

x

Examples for the case when f is continuous at x = a

22)2(248)321(

3191)3)1(2)1(()1()(lim

,

.139)32()(

,

1.),,3(3)(

1.9)32()(

1.,9)(

1.,32)(

39)32(lim

)2(

9

95

1

95

95

9

5

95

1

fxf

soand

xatcontisxxxxxffunctionThe

Thus

xatcontisitthusoncontisxxhfunctionrootThe

xatcontisxxxxgfunctionproducttheTherefore

xatcontisitthuseverywherecontisxxsfunctionThe

xatcontisitthuseverywherecontisxxxpfunctionpolynomialThe

xxxx

Example

x

x

Examples for the case when f is continuous at x= a

155515)5(53lim

5.53)(

:,5.

).2

5..1

:(

.5)(3)(

53lim

:)3(

5

5

fxx

Thus

xatcontisxxxf

thusxatcontaretheysoand

hGraph

xatcontishthatShow

Questions

everywherecontarexxhandxxg

xx

Example

x

x

2. The case 0/0

Suppose we want to find:

For the case when:

Then this is called the case 0/0. Caution: The limit is not equal 0/0. This is just a name that classifies the type of limits having such property.

)(

)(lim

xh

xgax

.0)(lim&)(lim arexhxgaxax

Examples for the case 0/0

8

3

32

12

)8(4

444

)4)(2(

42lim

)4)(2)(2(

)42)(2(lim

16

8lim

16lim08lim

:,0/016

8lim

:)1(

2

2

2

2

2

2

4

3

2

4

2

3

2

4

3

2

xx

xx

xxx

xxx

x

x

xx

becausecasetheisThisx

x

factoringbySolvingExample

x

x

x

xx

x

Examples for the case 0/0

30

1

)55(3

1

525)3(

1lim

525)3(lim

525)3(

25)25(lim

525

525.

)3(

525lim

)3(

525lim

)3(lim0525lim

:,0/0

)3(

525lim

)2(

0

00

00

00

0

xx

xxx

x

xxx

x

x

x

xx

x

xx

x

xxx

becausecasetheisThis

xx

x

methodconjugatethebygMultiplyinExample

x

xx

xx

xx

x

Examples for the case 0/0

existnotDoesx

xx

x

x

xThus

x

x

x

x

x

x

x

x

x

x

x

x

x

thatNotice

xx

becausecasetheisThis

Solutionx

xexistsifFind

ValuesAbsoluteInvolvingExample

x

xx

xxxx

xxxx

xx

xx

xx

x

153

102lim

153

102lim

153

102lim:

3

2

3

2lim

)5(3

)5(2lim

153

)102(lim

153

102lim

3

2

3

2lim

)5(3

)5(2lim

153

102lim

153

102lim

102

153lim0102lim

:,0/0

:153

102lim:

)3(

5

)5()5(

)5()5()5()5(

)5()5()5()5(

5;102

5;)102(

55

5

Question: Simplify the formula of f and graph it!

5;3

2

5;3

2

5;153

102

5;153

)102(153

102)(

x

x

xx

x

xx

xx

xxf

Questions

fGraphc

fofformulatheSimplifyb

xpotheatittheand

itlefttheitlefttheexistseitherifFinda

x

xfLet

fGraphc

fofformulatheSimplifyb

xpotheatittheand

itlefttheitlefttheexistseitherifFinda

x

xfLet

.

.

.0intlim

lim,lim,,.

3

62.2

.

.

.0intlim

lim,lim,,.

.1

3. The case ∞/ ∞

Suppose we want to find:

For the case when the limits of both functions f and g are infinite

Then this is called the case ∞/ ∞. Caution: The limit is not equal ∞/∞. This is just a name that classifies the type of limits having such property.

)(

)(lim

)(

)(lim

xh

xgOr

xh

xgxax

Limits at infinity

A function y=f(x) may approach a real number b as x increases or decreases with no bound.When this happens, we say that f has a limit at infinity, and that the line y=b is a horizontal asymptote for f.

Limit at infinity: The Case of Rational Functions

A rational function r(x) = p(x)/q(x) has a limit at infinity if the degree of p(x) is equal or less than the degree of q(x).

A rational function r(x) = p(x)/q(x) does not have a limit at infinity (but has rather infinite right and left limits) if the degree of p(x) is greater than the degree of q(x).

Example (1)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 9, is equal to the degree of the polynomial in the denominator, then

146

325)(

79

29

xx

xxxf

)(lim xfx

To show that, we follow the following steps:

6

5

)(lim9

9

inatordenomtheinxofcofficientThe

numeratortheinxofcofficientThexf

x

6

5

0)0(46

)0(3)0(25

1lim

1lim46lim

1lim3

1lim25lim

1146lim

3125lim

1146

3125

lim146

325lim)(lim

92

97

92

97

92

97

79

29

xx

xx

xx

xx

xx

xxxx

xxxf

xxx

xxx

x

x

xxx

Example (2)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 9, is less than the degree of the polynomial in the denominator, which is 12, then

146

325)(

712

29

xx

xxxf

)(lim xfx

To show that, we follow the following steps:

0)(lim

xfx

06

0

0)0(46

)0(3)0(2)0(5

1lim

1lim46lim

1lim3

1lim2

1lim5

1146lim

312

5lim

1146

312

5

lim146

325lim)(lim

125

12103

125

12103

125

12103

712

29

xx

xxx

xx

xxx

xx

xxxxx

xxxf

xxx

xxx

x

x

xxx

Example (3)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 9, then

146

325)(

79

212

xx

xxxf

)(lim xfx

They are infinite limits. To show that, we follow the following steps:

3

9

12

79

212

lim

6

5lim

146

325lim)(lim

xassamethearewhich

x

xassametheare

xx

xxxf

x

x

xx

.)(lim existnotdoxfx

Example (4)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 9, then

146

325)(

79

212

xx

xxxf

)(lim xfx

They are infinite limits. To show that, we follow the following steps:

)(lim

6

5lim

146

325lim)(lim

3

9

12

79

212

xassamethearewhich

x

xassametheare

xx

xxxf

x

x

xx

.)(lim existnotdoxfx

Example (5)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 8, then

146

325)(

78

212

xx

xxxf

)(lim xfx

They are infinite limits. To show that, we follow the following steps:

.)(lim existnotdoxfx

4

8

12

78

212

lim

6

5lim

146

325lim)(lim

xassamethearewhich

x

xassametheare

xx

xxxf

x

x

xx

Example (6)Let

Find

Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 8, then

146

325)(

78

212

xx

xxxf

)(lim xfx

They are infinite limits. To show that, we follow the following steps:

.)(lim existnotdoxfx

)(lim

6

5lim

146

325lim)(lim

4

8

12

78

212

xassamethearewhich

x

xassametheare

xx

xxxf

x

x

xx

Limits & Infinity

Problems Involving Roots

Introduction

We know that:

√x2 = |x|, which is equal x is x non-negative and equal to – x if x is negative

For if x = 2, then √(2)2 = √4 = 2 = |2|& if x = - 2, then √(-2)2 = √4 = -(-2) =|-2|

Example

0;16

0;3

0;16

00;)16(

0;3

00;16

0;3

0;)16(

0;3

2

2

29

29

29

292

292

916)(

:

916)(

:

xx

x

xx

xorxx

x

xorxx

x

xx

x

x

x

x

xx

xxf

asrewwrittenbecan

xxf

Let

Example

fofasymptoteshorizontaltheareyandylinestheThus

Whyx

x

x

x

xx

xx

xx

x

xx

x

xx

x

x

xxf

formulaoneintogetheritsmlithesefindwillWe

atandatfofitsmlithefindfrstWe

Solution

fofasymptoteshorizontaltheFindx

xxf

Let

xx

xx

xxx

xxxx

22

?2

2

4

02

0162

lim2lim

9lim16lim

22

916

lim)

22(

916

lim22

916

lim

22

916

lim22

)9

16(lim

22

916lim)(lim

,

:

22

916)(

:

2

222

22

22

2

2

4. The case of infinite limit

Infinite Limits

A function f may increases or decreases with no bound near certain values c for the independent variable x. When this happens, we say that f has an infinite limit, and that f has a vertical asymptote at x = c The line x=c is called a vertical asymptote for f.

Infinite Limits

A function has an infinite one-sided limit at a point x=c if at that point the considered one-sided limit of the denominator is zero and that of the numerator is not zero. The sign of the infinite limit is determined by the sign of both the numerator and the denominator at values close to the considered point x=c approached by the variable x (from the considered side).

Infinite Limits- The Case of Rational Functions

A rational function has an infinite one-sided limit at a point x=c if c a zero of the denominator but not of the numerator. The sign of the infinite limit is determined by the sign of both the numerator and the denominator at values close to the considered point x=c approached by the variable x (from the considered side).

Example (1)Let

Find

Solution:First x=0 is a zero of the denominator which is not a zero of the numerator.

xxf

1)(

)(lim.0

xfax

)(lim.0

xfbx

a. As x approaches 0 from the right, the numerator is always positive ( it is equal to 1) and the denominator approaches 0 while keeping positive; hence, the function increases with no bound. Thus:

)(lim0

xfx

The function has a vertical asymptote at x = 0, which is the line x = 0 (see the graph in the file on basic algebraic functions).

b. As x approaches 0 from the left, the numerator is always positive ( it is equal to 1) and the denominator approaches 0 while keeping negative; hence, the function decreases with no bound.

)(lim0

xfx

Example (2)Let

Find

Solution:First x=1 is a zero of the denominator which is not a zero of the numerator.

)(lim.1

xfax

)(lim.1

xfbx

1

5)(

x

xxf

)(lim1

xfx

The function has a vertical asymptote at x = 1, which is the line x = 1

)(lim1

xfx

a. As x approaches 1 from the right, the numerator approaches 6 (thus keeping positive), and the denominator approaches 0 while keeping positive; hence, the function increases with no bound. Thus,

b. As x approaches 1 from the left, the numerator approaches 6 (thus keeping positive), and the denominator approaches 0 while keeping negative; hence, the function decreases with no bound. The function has a vertical asymptote at x=1, which is the line x = 1. Thus:

Example (3)Let

Find

Solution:First x=3 is a zero of the denominator which is not a zero of the numerator.

)(lim.3

xfax

)(lim.3

xfbx

)3)(1(

)4)(1(

34

45)(

2

2

xx

xx

xx

xxxf

)(lim

)033,3(

,03

;14343

3

4lim

)3)(1(

)4)(1(lim

34

45lim

)(lim

3

3

3

2

2

3

3

xfThus

xsoandxhavewexasbecause

positivekeepingwhilexand

negativekeepingthusxxAs

x

x

xx

xxxx

xx

xf

x

x

x

x

x

)(lim

)033,3(

,03

;14343

3

4lim

)3)(1(

)4)(1(lim

34

45lim

)(lim

3

3

3

2

2

3

3

xfThus

xsoandxhavewexasbecause

negativekeepingwhilexand

negativekeepingthusxxAs

x

x

xx

xxxx

xx

xf

x

x

x

x

x

6. The case constant/∞Suppose we want to find:

For the case when:

In this case, no mater what the formulas of g and h are, we will always have:

Then this is called the case c/∞. Caution: The limit is not equal c/ ∞. This is just a name that classifies the type of limits having such property. This limit is always equal zero

)(

)(lim

xh

xgax

)(lim&)(lim xhRcxgaxax

0)(

)(lim

xh

xgax

Example on the case constant/∞

01

1lim

)(

)(lim

,

?)()(lim1)(

11lim)(lim1)(

:

:

1

1lim

,

:)1(

xxxh

xg

Thus

Whyxhxxxh

xgxg

haveWe

Solution

xx

Find

Example

xx

x

xx

x

6. Using the Squeeze Theorem

The Squeeze (Sandwich or Pinching)) Theorem

Suppose that we want to find the limit of a function f at a given point x=a and that the values of f on some interval containing this point (with the possible exception of that point) lie between the values of a couple of functions g and h whose limits at x=a are equal. The squeeze theorem says that in this case the limit of f at x=a will equal the limit of g and h at this point.

The Squeeze Theory

lxf

Then

xhlxg

dcawhere

adcxxhxfxg

Let

ax

axax

)(

:

)()(

&

),(

),(;)()()(

:

lim

limlim

Example (1)

1)(

,,

)4,0(;)(12,)4,0(1

&

1)1(&112)12(

:

:

)(

)4,0(;)(12

:

lim

limlim

lim

1

2

22

11

1

2

xf

theoremsqueezethebyThus

xxxfx

xx

haveWe

Soluion

xf

Find

xxxfx

Let

x

xx

x

Example (2)

7)(

,,

),0(;)(12,),0(4

&

771616)74(

,7916)94(

:

:

)(

),0[;74)(94

:

lim

lim

lim

lim

4

2

2

4

4

4

2

xf

theoremsqueezethebyThus

xxxfx

xx

x

haveWe

Soluion

xf

Find

xxxxfx

Let

x

x

x

x

Example (3)

0)1

sin(

,

)(0)(

'

0)5,5(;1

sin

)...(

?0)5,5(;11

sin1

:

:

:

)1

sin(

2

0

2

0

2

0

222

22

2

0

lim

limlim

lim

xx

theoremsqueeztheBy

xx

haveWe

xxx

xx

negativenonisxthatNotexbygMultiplyin

Whyxx

haveWe

Soluion

xx

Find

x

xx

x

Example (4)

0)2

cos(

,

)(0)(

'

0)5,5(;2

cos

)...(

?0)5,5(;12

cos1

:

:

:

)2

cos(

4

0

4

0

4

0

444

44

4

0

lim

limlim

lim

xx

theoremsqueezetheBy

xx

haveWe

xxx

xx

negativenonisxthatNotexbygMultiplyin

Whyxx

haveWe

Soluion

xx

Find

x

xx

x

Example (5)

07

cos7

,

707

'

}7{),(;77

cos77

}7{),(,7(

}7{),(;17

cos1

:

}7{),(;17

cos1

:

:

:

7cos7

3

7

77

3

3

3

7

lim

limlim

lim

xx

theoremsqueezetheBy

xx

haveWe

xxx

xx

onpositiveiswhichxbygMultiplyin

xx

soand

xx

haveWe

Soluion

xx

Find

x

xx

x

Example (6)

1)(

,

11112)112(

2)3,0(;)(112

)3,0(2

:

)(

2)3,0(;)(112

:

lim

limlim

lim

2

02

22

2

2

2

xf

theoremsqueezetheBy

eexx

and

xexfxand

Soluion

xf

Find

xexfx

Let

x

x

xx

x

x

x

2)( xexh112)( xxg

Example (7)

1)(

,

110coscos

0)2

,2

(;1)(cos

)2

,2

(0

:

)(

0)2

,2

(;1)(cos

:

lim

limlim

lim

0

00

0

xf

theoremsqueezetheBy

x

and

xxfxand

Soluion

xf

Find

xxfx

Let

x

xx

x

Example (8)

1sin

:),8(,

0)2

,2

(;1sin

cos

:

:

sin

lim

lim

0

0

x

x

atarriveweExampleinascontinuingThus

xx

xx

thatshownbecanIt

Soluion

x

x

Find

x

x

Example (9)

1sin

5

)5sin(

:

]0)5([

0,5,

5:

:

5

)5sin(

limlim

limlim

lim

05

55

5

t

t

x

x

Thus

xtSince

approachestapproachesxasThen

xtLet

Soluion

x

x

Find

tx

xx

x