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King Abdulaziz University

1.5 Limits Involving Infinity; Asymptotes

Dr. Hamed Al-Sulami

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c© 2008 [email protected] http://www.kau.edu.sa/hhaalsalmi

Prepared: December 29, 2008 Presented: December 29, 2008, 2008

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http://www.kau.edu.sa

mailto:[email protected]

http://www.kau.edu.sa/hhaalsalmi

2/18

Limits Involving Infinity; Asymptotes

Infinite Limits

Let f(x) = 1x , from the table below,

x > 0 f(x)

.1 10

.01 102

.001 103

.0001 104

.00001 105

x < 0 f(x)

−.1 −10

−.01 −102

−.001 −103

−.0001 −104

−.00001 −105

Look at the Figure to the right.We can seethat f(x) decreases (increases) without boundas x approaches 0 from the left(right). Wedescribe this behavior by writing

limx→0−

1x

= −∞ and limx→0+

1x

= ∞.

1000

2000

3000

4000

5000

-1000

-2000

-3000

-4000

-5000

1 2 3-1-2-3

x

y

f(x) = 1x

��

x < 0 →: -2.0 f(x) →: -250.0

x > 0 →:2.0 f(x) →: 250.0

��

x < 0 →: -1.97 f(x) →: -253.34

x > 0 →:1.97 f(x) →: 253.34

��

x < 0 →: -1.95 f(x) →: -256.78

x > 0 →:1.95 f(x) →: 256.78

��

x < 0 →: -1.92 f(x) →: -260.3

x > 0 →:1.92 f(x) →: 260.3

��

��

x < 0 →: -1.89 f(x) →: -263.93

x > 0 →:1.89 f(x) →: 263.93

��

��

x < 0 →: -1.87 f(x) →: -267.66

x > 0 →:1.87 f(x) →: 267.66

��

��

x < 0 →: -1.84 f(x) →: -271.49

x > 0 →:1.84 f(x) →: 271.49

��

��

x < 0 →: -1.82 f(x) →: -275.44

x > 0 →:1.82 f(x) →: 275.44

��

��

x < 0 →: -1.79 f(x) →: -279.5

x > 0 →:1.79 f(x) →: 279.5

��

��

x < 0 →: -1.76 f(x) →: -283.69

x > 0 →:1.76 f(x) →: 283.69

��

��

x < 0 →: -1.74 f(x) →: -288.0

x > 0 →:1.74 f(x) →: 288.0

��

��

x < 0 →: -1.71 f(x) →: -292.45

x > 0 →:1.71 f(x) →: 292.45

��

��

x < 0 →: -1.68 f(x) →: -297.03

x > 0 →:1.68 f(x) →: 297.03

��

��

x < 0 →: -1.66 f(x) →: -301.76

x > 0 →:1.66 f(x) →: 301.76

��

��

x < 0 →: -1.63 f(x) →: -306.64

x > 0 →:1.63 f(x) →: 306.64

��

��

x < 0 →: -1.6 f(x) →: -311.69

x > 0 →:1.6 f(x) →: 311.69

��

��

x < 0 →: -1.58 f(x) →: -316.9

x > 0 →:1.58 f(x) →: 316.9

��

��

x < 0 →: -1.55 f(x) →: -322.29

x > 0 →:1.55 f(x) →: 322.29

��

��

x < 0 →: -1.52 f(x) →: -327.87

x > 0 →:1.53 f(x) →: 327.87

��

��

x < 0 →: -1.5 f(x) →: -333.64

x > 0 →:1.5 f(x) →: 333.64

��

��

x < 0 →: -1.47 f(x) →: -339.62

x > 0 →:1.47 f(x) →: 339.62

��

��

x < 0 →: -1.45 f(x) →: -345.82

x > 0 →:1.45 f(x) →: 345.82

��

��

x < 0 →: -1.42 f(x) →: -352.25

x > 0 →:1.42 f(x) →: 352.25

��

��

x < 0 →: -1.39 f(x) →: -358.92

x > 0 →:1.39 f(x) →: 358.92

��

��

x < 0 →: -1.37 f(x) →: -365.85

x > 0 →:1.37 f(x) →: 365.85

��

��

x < 0 →: -1.34 f(x) →: -373.06

x > 0 →:1.34 f(x) →: 373.06

��

��

x < 0 →: -1.31 f(x) →: -380.55

x > 0 →:1.31 f(x) →: 380.55

��

��

x < 0 →: -1.29 f(x) →: -388.35

x > 0 →:1.29 f(x) →: 388.35

��

��

x < 0 →: -1.26 f(x) →: -396.48

x > 0 →:1.26 f(x) →: 396.48

��

��

x < 0 →: -1.23 f(x) →: -404.95

x > 0 →:1.23 f(x) →: 404.95

��

��

x < 0 →: -1.21 f(x) →: -413.79

x > 0 →:1.21 f(x) →: 413.79

��

��

x < 0 →: -1.18 f(x) →: -423.03

x > 0 →:1.18 f(x) →: 423.03

��

��

x < 0 →: -1.16 f(x) →: -432.69

x > 0 →:1.16 f(x) →: 432.69

��

��

x < 0 →: -1.13 f(x) →: -442.8

x > 0 →:1.13 f(x) →: 442.8

��

��

x < 0 →: -1.1 f(x) →: -453.4

x > 0 →:1.1 f(x) →: 453.4

��

��

x < 0 →: -1.08 f(x) →: -464.52

x > 0 →:1.08 f(x) →: 464.52

��

��

x < 0 →: -1.05 f(x) →: -476.19

x > 0 →:1.05 f(x) →: 476.19

��

��

x < 0 →: -1.02 f(x) →: -488.47

x > 0 →:1.02 f(x) →: 488.47

��

��

x < 0 →: -1.0 f(x) →: -501.39

x > 0 →:1.0 f(x) →: 501.39

��

��

x < 0 →: -0.97 f(x) →: -515.02

x > 0 →:0.97 f(x) →: 515.02

��

��

x < 0 →: -0.94 f(x) →: -529.41

x > 0 →:0.94 f(x) →: 529.41

��

��

x < 0 →: -0.92 f(x) →: -544.63

x > 0 →:0.92 f(x) →: 544.63

��

��

x < 0 →: -0.89 f(x) →: -560.75

x > 0 →:0.89 f(x) →: 560.75

��

��

x < 0 →: -0.87 f(x) →: -577.85

x > 0 →:0.87 f(x) →: 577.85

��

��

x < 0 →: -0.84 f(x) →: -596.03

x > 0 →:0.84 f(x) →: 596.03

��

��

x < 0 →: -0.81 f(x) →: -615.38

x > 0 →:0.81 f(x) →: 615.38

��

��

x < 0 →: -0.79 f(x) →: -636.04

x > 0 →:0.79 f(x) →: 636.04

��

��

x < 0 →: -0.76 f(x) →: -658.14

x > 0 →:0.76 f(x) →: 658.14

��

��

x < 0 →: -0.73 f(x) →: -681.82

x > 0 →:0.73 f(x) →: 681.82

��

��

x < 0 →: -0.71 f(x) →: -707.27

x > 0 →:0.71 f(x) →: 707.27

��

��

x < 0 →: -0.68 f(x) →: -734.69

x > 0 →:0.68 f(x) →: 734.69

��

��

x < 0 →: -0.65 f(x) →: -764.33

x > 0 →:0.65 f(x) →: 764.33

��

��

x < 0 →: -0.63 f(x) →: -796.46

x > 0 →:0.63 f(x) →: 796.46

��

��

x < 0 →: -0.6 f(x) →: -831.41

x > 0 →:0.6 f(x) →: 831.41

��

��

x < 0 →: -0.57 f(x) →: -869.57

x > 0 →:0.58 f(x) →: 869.57

��

��

x < 0 →: -0.55 f(x) →: -911.39

x > 0 →:0.55 f(x) →: 911.39

��

��

x < 0 →: -0.52 f(x) →: -957.45

x > 0 →:0.52 f(x) →: 957.45

��

��

x < 0 →: -0.5 f(x) →: -1008.4

x > 0 →:0.5 f(x) →: 1008.4

��

��

x < 0 →: -0.47 f(x) →: -1065.09

x > 0 →:0.47 f(x) →: 1065.09

��

��

x < 0 →: -0.44 f(x) →: -1128.53

x > 0 →:0.44 f(x) →: 1128.53

��

��

x < 0 →: -0.42 f(x) →: -1200.0

x > 0 →:0.42 f(x) →: 1200.0

��

��

x < 0 →: -0.39 f(x) →: -1281.14

x > 0 →:0.39 f(x) →: 1281.14

��

��

x < 0 →: -0.36 f(x) →: -1374.05

x > 0 →:0.36 f(x) →: 1374.05

��

��

x < 0 →: -0.34 f(x) →: -1481.48

x > 0 →:0.34 f(x) →: 1481.48

��

��

x < 0 →: -0.31 f(x) →: -1607.14

x > 0 →:0.31 f(x) →: 1607.14

��

��

x < 0 →: -0.28 f(x) →: -1756.1

x > 0 →:0.28 f(x) →: 1756.1

��

��

x < 0 →: -0.26 f(x) →: -1935.48

x > 0 →:0.26 f(x) →: 1935.48

��

��

x < 0 →: -0.23 f(x) →: -2155.69

x > 0 →:0.23 f(x) →: 2155.69

��

��

x < 0 →: -0.21 f(x) →: -2432.43

x > 0 →:0.21 f(x) →: 2432.43

��

��

x < 0 →: -0.18 f(x) →: -2790.7

x > 0 →:0.18 f(x) →: 2790.7

��

��

x < 0 →: -0.15 f(x) →: -3272.73

x > 0 →:0.15 f(x) →: 3272.73

��

��

x < 0 →: -0.13 f(x) →: -3956.04

x > 0 →:0.13 f(x) →: 3956.04

��

��

limx→0−

f(x) = −∞

limx→0+

f(x) = ∞

� � � � �

Limits Involving Infinity; Asymptotes 3/18

Note 1: The expressions limx→a

f(x) = ∞ and limx→a

f(x) = −∞ do not mean that the limit of

the function exists as x approaches a (it does not exist), but we use this to indicate that thefunction increase or decrease without bound near the number a. Another thing to be pointout is we say that f(x) approaches ∞ as x approaches a

(limx→a

f(x) = ∞)

if and only if f(x)

approaches ∞ as x approaches a from the right and the left(

limx→a+

f(x) = ∞ = limx→a−

f(x))

.

If the limit of f(x) from one side of a is ∞ and from the other side of a is −∞, then limx→a

f(x)

does not exist and we usually write limx→a

f(x) DNE.

!�� "�#�� $�� %& '(� ) limx→a

f(x) = −∞ � limx→a

f(x) = ∞ '* �� +(��

�� ,�� %& -�*�� +(�� /0 1�2�3� �� 4�� (�"�#�� 6 �� $��) a �� x.�8 %��9 :� ��

�� ;<

(limx→a

f(x) = ∞)

a �� x !�� ∞ �� !�� f(x) =�� 8 �>� ��

a �� ? 3�� @�� a �� x !�� ∞ �� !�� f(x) %&

.

(lim

x→a+f(x) = ∞ = lim

x→a−f(x)

)

�$A�� $�� ∞ �0 ��$A�� 8� �� a �� x !�� f(x) �� $�� B� � �C&

. limx→a

f(x) DNE �"�#�� 6 �� $�� %& =�� ;< −∞ �0 �>)�

� � � � �


Example 1.

Use the graph of the function f to find thefollowing limits if it exist. If the limit doesnot exist explain why?

1. limx→1

f(x)

2. limx→−1

f(x)

3. limx→0

f(x)1

2

3

4

5

6

−1

−2

−3

1 2 3 4−1−2−3−4−5

x

y

y = f(x)

Solution:

1. Since limx→1−

f(x) = ∞, and limx→1+

f(x) = −∞, then limx→1

f(x) DNE. ?

2. Since limx→−1−

f(x) = ∞ = limx→1+

f(x), then limx→1

f(x) = ∞. ?

3. Since limx→0−

f(x) = 3 = limx→0+

f(x), then limx→0

f(x) = 3. ?

�

� � � � �

1

2

3

4

5

6

-1

-2

-3

1 2 3 4-1-2-3-4-5

x

y

y = f(x)

��

1

2

3

4

5

6

-1

-2

-3

1 2 3 4-1-2-3-4-5

x

y

y = f(x)

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1

2

3

4

5

6

-1

-2

-3

1 2 3 4-1-2-3-4-5

x

y

y = f(x)

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Suppose that limx→a

f(x) = L �= 0, and limx→a

g(x) = 0. Then

limx→a+

f(x)g(x)

={

∞, if f(x), g(x) > 0 or f(x), g(x) < 0 for x > a;−∞, if f(x) > 0, g(x) < 0 or f(x) < 0, g(x) > 0 for x > a. ,

Similarly,

limx→a−

f(x)g(x)

={

∞, if f(x), g(x) > 0 or f(x), g(x) < 0 for x < a;−∞, if f(x) > 0, g(x) < 0 or f(x) < 0, g(x) > 0 for x < a. .

Now, if limx→a+

f(x)g(x)

= limx→a−

f(x)g(x)

, the we write limx→a

f(x)g(x)

= limx→a+

f(x)g(x)

= limx→a−

f(x)g(x)

.

If limx→a+

f(x)g(x)

�= limx→a−

f(x)g(x)

, the we write limx→a

f(x)g(x)

DNE. So to compute limx→a

f(x)g(x)

find

limx→a+

f(x)g(x)

and limx→a−

f(x)g(x)

. Now, if limx→a+

f(x)g(x)

and limx→a−

f(x)g(x)

are both ∞ or both −∞,

then limx→a

f(x)g(x)

equal the common limit. If limx→a+

f(x)g(x)

and limx→a−

f(x)g(x)

are different, then

limx→a

f(x)g(x)

DNE.

� � � � �


Example 2. Evaluate the following limits

1. limx→3+

x + 1

x2 − x − 62. lim

x→−1−x2 + 1

4x2 + 16x + 12

Solution:

1. limx→3+

x + 1

x2 − x − 6= lim

x→3+

x + 1

x2 − x − 6A direct substation gives I.F. 4/0. If x > 3

= limx→3+

x + 1

(x − 3)(x + 2)and near 3,then x + 1 > 0, x − 3 > 0, x + 2 > 0.

= limx→3+

+

x + 1+

(x − 3)+

(x + 2)

= ∞.?

2. limx→−1−

x2 + 1

4x2 + 16x + 12= lim

x→−1−x2 + 1

4x2 + 16x + 12A direct substation gives I.F. 2/0. If x < −1 and

= limx→−1−

x2 + 1

4(x + 1)(x + 3)near −1,then x2 + 1 > 0, x + 1 < 0, x + 3 > 0.

= limx→−1−

+

x2 + 1+

4−

(x + 1)+

(x + 3)

= −∞.?

�

� � � � �

1

2

-1

-2

-3

-4

-5

1 2-1-2-3-4-5

x

y

��

��

��

��

1

2

3

4

5

6

-1

-2

-3

1 2 3 4 5 6-1

x

y

y = f(x)

��

��

��

��

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Vertical Asymptote

Definition .1: [Vertical Asymptote]The line x = a is called vertical asymptote of the graph of y = f(x) if one of the followingcases happen: lim

x→a−f(x) = ±∞, lim

x→a+f(x) = ±∞.

Note 2: The vertical asymptote for a quotient function is likely to occur at the zeroesof the denominator. One thing we want to pint out here is that not every zeros of thedenominator is a vertical asymptote. Most of the students think that every zero ofthe denominator is a vertical asymptote which is not true.

Example 3. Find all vertical asymptotes of the graph of f(x) =x + 2

x2 − 4

Solution: Write f(x) =x + 2

(x − 2)(x + 2). The zeroes of the denominator are −2, 2. To chuck that

x = −2 is a vertical asymptote or not we take the limit at −2 from both sides.

limx→−2

f(x) = limx→−2

��(x + 2)

��(x + 2)(x − 2)= lim

x→−2

1

x − 2=

−1

4. Hence x = −2

?is not a vertical asymptote.

To chuck that x = 2 we take the limit limx→2+

f(x) = limx→2+

+

x + 2+

(x + 2)(x − 2)

= ∞. Hence x = 2?

is a

vertical asymptote. �

� � � � �

1

2

3

4

5

-1

-2

-3

-4

-5

-6

1 2 3 4 5-1-2-3-4-5-6�� x

y

y = f(x)

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Limits at Infinity

Let f(x) =4x2

x2 + 1, from the table be-

low,x > 0 f(x)

1 2

10 3.96

100 3.999

x < 0 f(x)

−1 2

−10 3.96

−100 3.999

Look at the Figure to the right.we cansee that the values of f(x) appear to ap-proach 4 as x increases without bound(x → ∞) or decreases without bound(x → −∞). We describe this behaviorby writing

limx→∞ f(x) = lim

x→∞4x2

x2 + 1= 4, and

limx→−∞ f(x) = lim

x→−∞4x2

x2 + 1= 4.

1

2

3

4

5

20 40 60 80 100 120 140-20-40-60-80-100-120-140

x

y

f(x) = 4x2

x2+1

��

x < 0 →: 0.0 f(x) →: 0.0 x > 0 →:0.0 f(x) →: 0.0

��

x < 0 →: -1.94 f(x) →: 0.04 x > 0 →:1.94 f(x) →: 0.04

��

x < 0 →: -3.89 f(x) →: 0.15 x > 0 →:3.89 f(x) →: 0.15

��

x < 0 →: -5.83 f(x) →: 0.31 x > 0 →:5.83 f(x) →: 0.31

��

x < 0 →: -7.78 f(x) →: 0.53 x > 0 →:7.78 f(x) →: 0.53

��

x < 0 →: -9.72 f(x) →: 0.76 x > 0 →:9.72 f(x) →: 0.76

��

x < 0 →: -11.67 f(x) →: 1.02 x > 0 →:11.67 f(x) →: 1.02

��

x < 0 →: -13.61 f(x) →: 1.27 x > 0 →:13.61 f(x) →: 1.27

��

x < 0 →: -15.56 f(x) →: 1.51 x > 0 →:15.56 f(x) →: 1.51

��

x < 0 →: -17.5 f(x) →: 1.73 x > 0 →:17.5 f(x) →: 1.73

��

x < 0 →: -19.44 f(x) →: 1.94 x > 0 →:19.44 f(x) →: 1.94

��

x < 0 →: -21.39 f(x) →: 2.13 x > 0 →:21.39 f(x) →: 2.13

��

x < 0 →: -23.33 f(x) →: 2.31 x > 0 →:23.33 f(x) →: 2.31

��

x < 0 →: -25.28 f(x) →: 2.46 x > 0 →:25.28 f(x) →: 2.46

��

x < 0 →: -27.22 f(x) →: 2.6 x > 0 →:27.22 f(x) →: 2.6

��

x < 0 →: -29.17 f(x) →: 2.72 x > 0 →:29.17 f(x) →: 2.72

��

x < 0 →: -31.11 f(x) →: 2.83 x > 0 →:31.11 f(x) →: 2.83

��

x < 0 →: -33.06 f(x) →: 2.93 x > 0 →:33.06 f(x) →: 2.93

��

x < 0 →: -35.0 f(x) →: 3.02 x > 0 →:35.0 f(x) →: 3.02

��

x < 0 →: -36.94 f(x) →: 3.09 x > 0 →:36.94 f(x) →: 3.09

��

x < 0 →: -38.89 f(x) →: 3.16 x > 0 →:38.89 f(x) →: 3.16

��

x < 0 →: -40.83 f(x) →: 3.23 x > 0 →:40.83 f(x) →: 3.23

��

x < 0 →: -42.78 f(x) →: 3.28 x > 0 →:42.78 f(x) →: 3.28

��

x < 0 →: -44.72 f(x) →: 3.33 x > 0 →:44.72 f(x) →: 3.33

��

x < 0 →: -46.67 f(x) →: 3.38 x > 0 →:46.67 f(x) →: 3.38

��

x < 0 →: -48.61 f(x) →: 3.42 x > 0 →:48.61 f(x) →: 3.42

��

x < 0 →: -50.56 f(x) →: 3.46 x > 0 →:50.56 f(x) →: 3.46

��

x < 0 →: -52.5 f(x) →: 3.49 x > 0 →:52.5 f(x) →: 3.49

��

x < 0 →: -54.44 f(x) →: 3.52 x > 0 →:54.44 f(x) →: 3.52

��

x < 0 →: -56.39 f(x) →: 3.55 x > 0 →:56.39 f(x) →: 3.55

��

x < 0 →: -58.33 f(x) →: 3.58 x > 0 →:58.33 f(x) →: 3.58

��

x < 0 →: -60.28 f(x) →: 3.6 x > 0 →:60.28 f(x) →: 3.6

��

x < 0 →: -62.22 f(x) →: 3.63 x > 0 →:62.22 f(x) →: 3.63

��

x < 0 →: -64.17 f(x) →: 3.65 x > 0 →:64.17 f(x) →: 3.65

��

x < 0 →: -66.11 f(x) →: 3.66 x > 0 →:66.11 f(x) →: 3.66

��

x < 0 →: -68.06 f(x) →: 3.68 x > 0 →:68.06 f(x) →: 3.68

��

x < 0 →: -70.0 f(x) →: 3.7 x > 0 →:70.0 f(x) →: 3.7

��

x < 0 →: -71.94 f(x) →: 3.71 x > 0 →:71.94 f(x) →: 3.71

��

x < 0 →: -73.89 f(x) →: 3.73 x > 0 →:73.89 f(x) →: 3.73

��

x < 0 →: -75.83 f(x) →: 3.74 x > 0 →:75.83 f(x) →: 3.74

��

x < 0 →: -77.78 f(x) →: 3.75 x > 0 →:77.78 f(x) →: 3.75

��

x < 0 →: -79.72 f(x) →: 3.76 x > 0 →:79.72 f(x) →: 3.76

��

x < 0 →: -81.67 f(x) →: 3.77 x > 0 →:81.67 f(x) →: 3.77

��

x < 0 →: -83.61 f(x) →: 3.78 x > 0 →:83.61 f(x) →: 3.78

��

x < 0 →: -85.56 f(x) →: 3.79 x > 0 →:85.56 f(x) →: 3.79

��

x < 0 →: -87.5 f(x) →: 3.8 x > 0 →:87.5 f(x) →: 3.8

��

x < 0 →: -89.44 f(x) →: 3.81 x > 0 →:89.44 f(x) →: 3.81

��

x < 0 →: -91.39 f(x) →: 3.82 x > 0 →:91.39 f(x) →: 3.82

��

x < 0 →: -93.33 f(x) →: 3.82 x > 0 →:93.33 f(x) →: 3.82

��

x < 0 →: -95.28 f(x) →: 3.83 x > 0 →:95.28 f(x) →: 3.83

��

x < 0 →: -97.22 f(x) →: 3.84 x > 0 →:97.22 f(x) →: 3.84

��

x < 0 →: -99.17 f(x) →: 3.84 x > 0 →:99.17 f(x) →: 3.84

��

x < 0 →: -101.11 f(x) →: 3.85 x > 0 →:101.11f(x) →: 3.85

��

x < 0 →: -103.06 f(x) →: 3.85 x > 0 →:103.06f(x) →: 3.85

��

x < 0 →: -105.0 f(x) →: 3.86 x > 0 →:105.0 f(x) →: 3.86

��

x < 0 →: -106.94 f(x) →: 3.86 x > 0 →:106.94f(x) →: 3.86

��

x < 0 →: -108.89 f(x) →: 3.87 x > 0 →:108.89f(x) →: 3.87

��

x < 0 →: -110.83 f(x) →: 3.87 x > 0 →:110.83f(x) →: 3.87

��

x < 0 →: -112.78 f(x) →: 3.88 x > 0 →:112.78f(x) →: 3.88

��

x < 0 →: -114.72 f(x) →: 3.88 x > 0 →:114.72f(x) →: 3.88

��

x < 0 →: -116.67 f(x) →: 3.89 x > 0 →:116.67f(x) →: 3.89

��

x < 0 →: -118.61 f(x) →: 3.89 x > 0 →:118.61f(x) →: 3.89

��

x < 0 →: -120.56 f(x) →: 3.89 x > 0 →:120.56f(x) →: 3.89

��

x < 0 →: -122.5 f(x) →: 3.9 x > 0 →:122.5 f(x) →: 3.9

��

x < 0 →: -124.44 f(x) →: 3.9 x > 0 →:124.44f(x) →: 3.9

��

x < 0 →: -126.39 f(x) →: 3.9 x > 0 →:126.39f(x) →: 3.9

��

x < 0 →: -128.33 f(x) →: 3.91 x > 0 →:128.33f(x) →: 3.91

��

x < 0 →: -130.28 f(x) →: 3.91 x > 0 →:130.28f(x) →: 3.91

��

x < 0 →: -132.22 f(x) →: 3.91 x > 0 →:132.22f(x) →: 3.91

��

x < 0 →: -134.17 f(x) →: 3.91 x > 0 →:134.17f(x) →: 3.91

��

x < 0 →: -136.11 f(x) →: 3.92 x > 0 →:136.11f(x) →: 3.92

��

x < 0 →: -138.06 f(x) →: 3.92 x > 0 →:138.06f(x) →: 3.92

�� lim

x→−∞ f(x) = 4 limx→∞ f(x) = 4

� � � � �


Theorem .1: [Properties of Limits as x → ±∞]Let c be real numbers, n be a positive integer, k be a positive rational number. Supposethat lim

x→±∞ f(x) = L and limx→±∞ g(x) = M. Then

1. Sum-Difference: limx→±∞[f(x) ± g(x)] = L ± M

2. Product: limx→±∞[f(x)g(x)] = LM

3. Quotient: limx→±∞[

f(x)g(x)

] =L

Mprovided M �= 0.

4. Constant multiple: limx→±∞[cf(x)] = cL

5. Power: limx→±∞[f(x)]n = Ln

6. Root: limx→±∞

n√

f(x) = n√

L provided L ≥ 0 if n is even

7. Constant limit limx→±∞ c = c

8. Reciprocal of Power limx→∞

c

xk= 0

9. Reciprocal of Power limx→−∞

c

xk= 0 provided xk is always defined

� � � � �


Infinite Limits At Infinity

Theorem .2: [Infinite Limits At Infinity]Let n be a positive integer, then

1. limx→∞xn = ∞.

2. limx→−∞xn =

{∞, if n is even ;−∞, if n is odd.

3. If p(x) = anxn + an−1xn−1 + ... + a2x

2 + a1x + a0, then limx→±∞ p(x) = lim

x→±∞anxn.

4. If limx→±∞ f(x) = ∞ and lim

x→±∞ g(x) = ∞, then limx→±∞ f(x)g(x) = ∞

5. If limx→±∞ f(x) = −∞ and lim

x→±∞ g(x) = ∞, then limx→±∞ f(x)g(x) = −∞

6. If limx→±∞ f(x) = −∞ and lim

x→±∞ g(x) = −∞, then limx→±∞ f(x)g(x) = ∞

Example 4. Find the following limits

1. limx→∞(−7x4 − 3x2 + 7) 2. lim

x→−∞3√

x − 1

Solution: 1. Since limx→∞

(−7x4) = −7 limx→∞

x4 = −7 · ∞ = −∞, then limx→∞

(−7x4 − 3x2 + 7) =

limx→∞

(−7x4) = −∞.

2. Since limx→−∞

(x − 1) = −∞, then limx→−∞

3√

x − 1 = −∞. �

� � � � �


Note 3: As we have seen that 0/0 is indeterminate form( I.F.). The following expressions∞∞ ,−∞

∞ ,∞ −∞ are all indeterminate forms (I.F.). When you get an indeterminate formsuch as ∞

∞ the best way to find the limit is to divide each term in the numerator and eachterm in the denominator by the highest power in the denominator and use laws (8-9), andwhen you get ∞−∞ indeterminate form,try to make a common denominator or rationalizethe numerator and use laws (8-9). The following are not indeterminate form. For any realnumber a. ∞± a = ∞ a · ∞ = ∞ if a > 0, a · ∞ = −∞ if a < 0. ∞ + ∞ = ∞, 1/∞ = 0.

.∞∞ ,−∞

∞ ,∞−∞, 0 ·∞ �� 9 (�� $�� D�E� �<�(� ��6 ��@� 0/0 %;< F 9 G �&? @�

�@3� �0 �� $�� " A�H ��I �E<� %;<∞∞ J�� <�(� ��6 ��@� �� =��

. � � $�� 1��2�G� KL 1 @�� '< �� +�� 1 @�� M3+�� '< �8 ��

�� @�� 8�� =� 8 ∞−∞ J�� <�(� ��6 ��@� �� =��

. � � $�� 1��2�G� KL �� N<�@� 9 !�E��

%;< '�8 "�� a % � �C& .�<�(� ��6 � �@� B3�� @4��

∞± a = ∞

.a < 0 �CO a · ∞ = −∞ � a > 0, �CO a · ∞ = ∞∞ + ∞ = ∞,

.1/∞ = 0

� � � � �


Example 5. Evaluate the each of following limit if it exist

1. limx→∞

2x + 1x2 + 7

2. limx→∞

2x2 + x + 2x2 + 2

Solution:

1. Since limx→∞

2x + 1 = ∞ = limx→∞

x2 + 7 we have I.F. type ∞/∞. Divide each term in the

numerator and each term in the denominator by the highest power in the denominator.

limx→∞

2x + 1

x2 + 7= lim

x→∞

2x + 1

x2

x2 + 7

x2

= limx→∞

2x

x2+

1

x2

x2

x2+

7

x2

= limx→∞

2

x+

1

x2

1 +7

x2

=0 + 0

1 + 0=

0

1= 0.

?

2. Divide the numerator and the denominator by x2

limx→∞

2x2 + x + 2

x2 + 2= lim

x→∞

2x2 + x + 2

x2

x2 + 2

x2

= limx→∞

2 +1

x+

2

x2

1 +2

x2

=2

1= 2.

?

�

� � � � �

1

2

-11 2 3 4 5 6 7 8 9 10-1

x

yy = 2x+1

x2+7

1

2

-11 2 3 4 5 6 7 8 9 10-1

x

yy = 2x+1

x2+7


Example 6. Evaluate the following limit if it exist limx→−∞

3x + 1√x2 + 1

.

Solution: For x < 0, we have −x = |x| =√

x2 hence x = −√

x2. Now, divide the numerator andthe denominator by x

limx→−∞

3x + 1√x2 + 1

= limx→−∞

3x + 1

x√x2 + 1

x

= limx→−∞

3x + 1

x√x2 + 1

−√

x2

x → −∞ ⇒ x < 0, x = −√

x2

= limx→−∞

3 +1

x

−√

x2 + 1

x2

= limx→−∞

3 +1

x

−√

1 +1

x2

=3 + 0

−√1 + 0

= −3.?

�

� � � � �

-1

-2

-3

-4

-1-2-3-4-5-6-7-8-9-10

x

yy = 3x+1√

x2+1


Example 7. Evaluate the each of following limit if it exist limx→∞

√x2 + x + 2

2 − x

Solution: For x > 0, we have x = |x| =√

x2 hence x =√

x2. Now, divide the numerator and thedenominator by x

limx→∞

√x2 + x + 2

2 − x= lim

x→∞

√x2 + x + 2

x2 − x

x

= limx→∞

√x2 + x + 2√

x2

2 − x

x

x → ∞ ⇒ x > 0, x =√

x2

= limx→∞

√x2 + x + 2

x2

2

x− 1

= limx→∞

√1 +

1

x+

2

x2

2

x− 1

=

√1 + 0 + 0

0 − 1= −1.

?

�

� � � � �

1

2

-1

-2

-3

-4

1 2 3 4 5 6 7 8 9 10 11 12-1

x

yy =

√x2+x+22−x


Definition .2: [Horizontal Asymptote]The line y = L is called a horizontal asymptote of the graph y = f(x) if one of thefollowing cases happen: lim

x→∞ f(x) = L or limx→−∞ f(x) = L

Note 4: From the definition we can see thatthe graph of a function can have at most twohorizontal asymptotes- one to right and one tothe left.

L1

L2

x

y

Example 8. Find the horizontal asymptotes for f(x) =3x2 + 12x − 4x2

Solution: To find the horizontal asymptote we take the limit as x → ±∞ Note that both thenumerator and the denominator x → ±∞, as x → ±∞. To find this limit divided both thenumerator and the denominator by the highest power of x in the denominator which is x2. So,

limx→±∞

3x2 + 1

2x − 4x2= lim

x→±∞

3x2 + 1

x2

2x − 4x2

x2

= limx→±∞

3 +1

x2

2

x− 4

=3 + 0

0 − 4=

−3

4.

Therefore y =−3

4?

is a horizontal asymptote. �

� � � � �

4

8

12

-4

-8

-12

-16

1 2 3-1-2-3-4y = −3/4 x

y

y = 3x2+12x−4x2


Example 9. Find the horizontal asymptotes for p(x) =3x − 1√

x2 + x + 1.

Solution: To find the horizontal asymptote, take the limit of the function as x → ±∞.

limx→−∞

3x − 1√x2 + x + 1

= limx→−∞

3x − 1

x√x2 + x + 1

x

= limx→−∞

3x − 1

x√x2 + x + 1

−√x2

x → −∞ ⇒ x < 0, x = −√

x2

= limx→−∞

3 − 1

x

−√

x2 + x + 1

x2

= limx→−∞

3 − 1

x

−√

1 +1

x+

1

x2

= −3.

Similarly, we have

limx→∞

3x − 1√x2 + x + 1

= limx→∞

3x − 1

x√x2 + x + 1

x

= limx→∞

3x − 1

x√x2 + x + 1√

x2

x → ∞ ⇒ x > 0, x =√

x2

= limx→∞

3 − 1

x√x2 + x + 1

x2

= limx→∞

3 − 1

x√1 +

1

x+

1

x2

=3 + 0√1 + 0

=3

1= 3.

Therefore y = ±3?

are horizontal asymptotes. �

� � � � �

1

2

3

4

5

-1

-2

-3

-4

-5

-6

1 2 3 4 5 6 7-1-2-3-4-5-6-7-8

x

y

y = 3x−1√x2+x+1


Example 10. Find the limit limx→∞[

√x2 − 3x + 1 − x].

Solution: A direct substation gives I.F. ∞−∞.

limx→∞

[√

x2 − 3x + 1 − x] = limx→∞

[√

x2 − 3x + 1 − x] ·√

x2 − 3x + 1 + x√x2 − 3x + 1 + x

Use (a − b)(a + b) = a2 − b2

= limx→∞

(x2 − 3x + 1) − x2

√x2 − 3x + 1 + x

Simplify divide by x

= limx→∞

−3x + 1

x√x2−3x+1+x

x

x → ∞ ⇒ x > 0, x =√

x2

= limx→∞

−3 +1

x√x2−3x+1

x+ 1

= limx→∞

−3 +1

x√x2−3x+1

x2 + 1

= limx→∞

−3 +1

x√1 − 3

x+ 1

x2 + 1=

−3 + 0√1 − 0 + 0 + 1

=−3

2.?

�

� � � � �

1

-1

-2

-3

-4

1 2 3 4 5 6 7 8 9-1

x

y y = [√

x2 − 3x + 1 − x]


Theorem .3: [The Squeeze Theorem ]If f(x) ≤ g(x) ≤ h(x) for all x in an open in-terval containing a, except possibly at a itselfand iflimx→a

f(x) = L = limx→a

h(x), then limx→a

g(x) = L.a

fg

h

L

x

y

Example 11. If 4 − x2 ≤ f(x) ≤ (3 +√

1 + x), find limx→0

f(x).

Solution: Since limx→0

(4−x2) = 4 = limx→0

(3+√

1 + x), then by The Squeeze Theorem we have

limx→0

f(x) = 4. �

Example 12. limx→∞

cosx

x

Solution: Since −1 ≤ cosx ≤ 1, then if x > 0 we have−1x

≤ cosx

x≤ 1

x. Now, since

limx→∞

−1x

= 0 = limx→∞

1x

, then by the Squeeze Theorem, we have limx→∞

cosx

x= 0. �

� � � � �