[PPT] - d Limits at infinity and infinite limits i E 2 Lectures a l l PowerPoint Presentation

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Section 2.6 Limits at infinity and infinite limits 2 Lectures

Dr. Abdulla Eid

College of Science

MATHS 101: Calculus I

Dr. Abdulla Eid (University of Bahrain)

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1 Finite limits as x → ±∞. 2 Horizontal Asympotes. 3 Infinite limits. 4 Vertical Asymptotes.

Dr. Abdulla Eid (University of Bahrain)

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Motivation Example

Consider the function f (x) = 1

x . The graph of the function is

x y Question: What happen if x is sufficiently large number (i.e., x approaches ∞) ? In other words, what is limx→∞ 1

x ?

From the graph we can easily see that lim

x→∞

1 x = 0 and lim

x→−∞

1 x = 0

Dr. Abdulla Eid (University of Bahrain)

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Continue...

Arithmetic at infinity:

1

∞ + ∞ = ∞.

2 k · ∞ = ∞ (k > 0). 3 k · ∞ = −∞ (k < 0). 4

1 ±∞ = 0.

5

∞ ∞ = ?. (Calculus 1).

6

∞ − ∞ =?, 0 · ∞ =?, 1∞ =?. (Calculus 2)

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Finding the limit of a rational function

To find the limit limx→±∞

f (x) g(x), we have

1 Substitute directly by x = ±∞ in f (x)

g(x). If you get a real number or

±∞, then that is the limit.

2 If you get undefined values such as 0

0 or ∞ ∞, we take the highest power

f x in the numerator and the highest power of x in the denominator

as common factor and we proceed.

Dr. Abdulla Eid (University of Bahrain)

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Example 1

Find lim

x→∞

3x2 − x − 2 5x2 + 4x + 1 Solution: Direct substitution gives 3(∞)2 − (∞) − 2 5(∞)2 + 4(∞) + 1 undefined! lim

x→∞

3x2 − x − 2 5x2 + 4x + 1 = lim

x→∞

3x2 − x − 2 5x2 + 4x + 1 = lim

x→∞

x2 3 − 1

x − 2 x2

x2

5 + 4

x + 1 x2

= lim

x→∞

3 − 1

x − 2 x2

5 + 4

x + 1 x2

= (3 − 0 − 0)

(5 + 0 + 0) = 3 5

Dr. Abdulla Eid (University of Bahrain)

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Exercise 2

Find lim

x→∞

5x2 − 2x + 1 9x2

Dr. Abdulla Eid (University of Bahrain)

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Example 3

Find lim

x→∞

3x + 7 x2 − 2 Solution: Direct substitution gives 3∞ + 7 (∞)2 − 2 = ∞ ∞ undefined! lim

x→∞

3x + 7 x2 − 2 = lim

x→∞

3x + 7 x2 − 2 = lim

x→∞

x

3 + 7

x

x2

1 − 2

x2

= lim

x→∞

3 + 7

x

x
1 − 2

x2

=

(3 + 0) ∞(1 − 0) = 0

Dr. Abdulla Eid (University of Bahrain)

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Example 4

Find lim

x→∞

x3 − 8 2x2 + 1 Solution: Direct substitution gives (∞)3 − 8 2(∞)2 + 1 = ∞ ∞ undefined! lim

x→∞

x3 − 8 2x2 + 1 = lim

x→∞

x3 − 8 2x2 + 1 = lim

x→∞

x3 1 − 8

x3

x2

2 + 1

x2

= lim

x→∞

x

1 − 8

x3

2 + 1

x2

= ∞
Dr. Abdulla Eid (University of Bahrain)

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Exercise 5

Find lim

x→∞

5x3 − 2x + 1 9x8 + 8

Dr. Abdulla Eid (University of Bahrain)

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Example 6

Find lim

x→∞

√ 3x2 + 1 3x − 5 Solution: Direct substitution gives lim

x→∞

3(∞)2 + 1

3(∞) − 5 = ∞ ∞ undefined! lim

x→∞

√ 3x2 + 1 3x − 5 = lim

x→∞

√ 3x2 + 1 3x − 5 = lim

x→∞

x2

3 + 1

x2

x
3 − 5

x

= lim

x→∞

√ x2

3 + 1

x2

x
3 − 5

x

= lim

x→∞

|x|

3 + 1

x2

x
3 − 5

x

= lim

x→∞

x

3 + 1

x2

x
3 − 5

x

= lim

x→∞

3 + 1

x2

3 − 5

= √ 3 3

Dr. Abdulla Eid (University of Bahrain)

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Exercise 7

Find lim

x→−∞

√ 3x2 + 1 3x − 5 Solution: Direct substitution gives lim

x→−∞

3(−∞)2 + 1

3(−∞) − 5 = ∞ −∞ undefined! lim

x→−∞

√ 3x2 + 1 3x − 5 = lim

x→∞

√ 3x2 + 1 3x − 5 = lim

x→−∞

x2

3 + 1

x2

x
3 − 5

x

=

lim

x→−∞

√ x2

3 + 1

x2

x
3 − 5

x

=

lim

x→−∞

|x|

3 + 1

x2

x
3 − 5

x

=

lim

x→−∞

−x

3 + 1

x2

x
3 − 5

x

=

lim

x→−∞

−

3 + 1

x2

3 − 5

x

= −

√ 3 3

Dr. Abdulla Eid (University of Bahrain)

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Multiplying by the conjugate

Example 8

Find lim

x→∞

x2 + 1 − x
Solution: Direct substitution gives
(∞)2 + 1 − x
= ∞ − ∞

undefined! lim

x→∞

x2 + 1 − x
= lim

x→∞ lim x→∞

x2 + 1 − x
·

√ x2 + 1 + x

√

x2 + 1 + x

= lim

x→∞

x2 + 1 − x2 √ x2 + 1 + x = lim

x→∞

1 √ x2 + 1 + x

= 0
Dr. Abdulla Eid (University of Bahrain)

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Exercise 9

Find lim

x→∞

x −
x2 + 16
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2 - Horizontal Asymptotes

Motivational Example: Consider the function f (x) = x2−1

x2+1. Then we have

lim

x→∞ f (x) = 1

and lim

x→∞ f (x) = 1

In this case, the line y = 1 is called a horizontal asymptote. x y

Dr. Abdulla Eid (University of Bahrain)

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Definition 10

The line y = L is called a horizontal asymptote of the curve y = f (x) if either lim

x→∞ f (x) = L

and lim

x→−∞ f (x) = L

Example 11

Find the horizontal asymptote of the function f (x) = x − 9 √ 4x2 + 3x + 2 we need to find both limx→∞ f (x) and limx→−∞ f (x)

Dr. Abdulla Eid (University of Bahrain)

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lim

x→∞

x − 9 √ 4x2 + 3x + 2 = lim

x→∞

x − 9 √ 3x2 + 3x + 1 = lim

x→∞

x

1 − 9

x

x2

4 + 3

x + 2 x2

= lim

x→∞

x

1 − 9

x

√

x2

4 + 3

x + 2 x2

= lim

x→∞

x

1 − 9

x

|x|
4 + 3

x + 2 x2

= lim

x→∞

x

1 − 9

x

x
4 + 3

x + 2 x2

= lim

x→∞

1 − 9

x

4 + 3

x + 2 x2

= 1 2 Hence y = 1 2 is a horizontal asymptote. Now we compute limx→−∞ f (x) to get limx→−∞ f (x) = −1

2 and so we have y = −1

2 is also a horizontal asymptote.

Dr. Abdulla Eid (University of Bahrain)

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Motivation Example

Consider the function f (x) = 1

x . The graph of the function is

x y Question: What is limx→0+ 1

x and limx→0− 1 x ? From the graph we can

easily see that lim

x→0+

1 x = ∞ 1 0+ = ∞

and

lim

x→0−

1 x = ∞ 1 0+ = ∞

Dr. Abdulla Eid (University of Bahrain)

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Example 12

Find lim

x→1+

3 x − 1 Solution: Direct substitution gives 3 undefined! So we need to find whether it is 0+ or 0−. lim

x→1+

3 x − 1 = 3 0+ = ∞

Dr. Abdulla Eid (University of Bahrain)

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Exercise 13

Find lim

x→1−

3 x − 1 Solution: Direct substitution gives 3 undefined! So we need to find whether it is 0+ or 0−. lim

x→1−

3 x − 1 = 3 0− = −∞

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Example 14

Find lim

x→−1+

−2 x + 1 Solution: Direct substitution gives −2 undefined! So we need to find whether it is 0+ or 0−. lim

x→−1+

−2 x + 1 = −2 0+ = −2 · ∞ = −∞

Dr. Abdulla Eid (University of Bahrain)

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Exercise 15

Find lim

x→2+

3 2 − x Solution: Direct substitution gives 3 undefined! So we need to find whether it is 0+ or 0−. lim

x→2+

3 2 − x = 3 0− = −∞

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Example 16

Find lim

x→4−

2x x2 − 16 Solution: Direct substitution gives 8 undefined! So we need to find whether it is 0+ or 0−. lim

x→4−

2x x2 − 16 = 8 0− = −∞

Dr. Abdulla Eid (University of Bahrain)

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Example 17

Find lim

x→2+

x − 2 x2 − 4x + 4 Solution: Direct substitution gives undefined! So we need to factor first using the methods of Section 2.2. lim

x→2+

x − 2 x2 − 4 = lim

x→2+

(x − 2) (x − 2)(x − 2) = lim

x→2+

1 x − 2 So we need to find whether it is 0+ or 0−. lim

x→2+

1 x − 2 = 1 0+ = ∞

Dr. Abdulla Eid (University of Bahrain)

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Exercise 18

Find lim

x→3

3x x2 − 9 Solution: Direct substitution gives 9 undefined! So we need to find whether it is 0+ or 0− and for that we find the right and the left limits. lim

x→3+

3x x2 − 9 = 9 0+ = ∞ lim

x→3−

3x x2 − 9 = 9 0− = − ∞ Since limx→3+

3x x2−9= limx→3− 3x x2−9, we have

lim

x→3

3x x2 − 9 Does Not Exist

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Example 19

Find lim

x→1

x2 − 1 (x − 1)2 Solution: Direct substitution gives 0 undefined! So we need to factor first using the methods of Section 2.2. lim

x→1

x2 − 1 (x − 1)2 = lim

x→1

(x − 1)(x + 1) (x − 1)(x − 1) = lim

x→1

x + 1 x − 1 So we need to find whether it is 0+ or 0− and for that we find the right and the left limits. lim

x→1+

x + 1 x − 1 = 2 0+ = ∞ lim

x→1−

x + 1 x − 1 = 2 0− = − ∞ Since limx→1+

x2−1 (x−1)2 = limx→1− x2−1 (x−1)2 , we have

lim

x→1

x2 − 1 (x − 1)2 Does Not Exist

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4 - Vertical Asymptotes

Motivational Example: Consider the function f (x) = x+3

x−2. Then we have

lim

x→2+ f (x) = ∞

and lim

x→2− f (x) = − ∞

In this case, the line x = 2 is called a vertical asymptote. x y

Dr. Abdulla Eid (University of Bahrain)

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Definition 20

The line x = a is called a vertical asymptote of the curve y = f (x) if either lim

x→a+ f (x) = ±∞

r

lim

x→a− f (x) = ±∞

To find the vertical asymptote for a rational function, we need to cancel any common factor first and we find where the denominator is zero.

Example 21

Find the vertical asymptote of the function f (x) = −8 x2 − 4 x2 − 4 = 0 → x = −2, x = 2. Since none of these is a zero for the numerator, then both are vertical asymptote.

Dr. Abdulla Eid (University of Bahrain)

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