Section 3.5 – Limits at Infinity. Vertical Asymptotes and Limits When we investigated infinite limits and vertical asymptotes, we let x approach a number.

Section 3.5 – Limits at Infinity

Vertical Asymptotes and LimitsWhen we investigated infinite limits and vertical

asymptotes, we let x approach a number. The result was that the values of y became arbitrarily large (positive or negative).

3

limx

f x

White Board Challenge

Analytically find the vertical asymptote(s) of:

22 6

9x

xh x

3x

Horizontal Asymptotes and LimitsWhen we investigate infinite limits and horizontal

asymptotes, we will let x become arbitrarily large (positive or negative) and see what happens to y. This will be referred to as the end behavior.

lim 2x

f x

End BehaviorLet f be a function defined on some interval (a,∞). Then

means that the values of f(x) get closer to L as x increases.

limx

f x L

lim 3x

f x

End BehaviorLet f be a function defined on some interval (-∞, a). Then

means that the values of f(x) get closer to L as x decreases.

limx

f x L

lim 3x

f x

End BehaviorLet f be a function defined on some interval (a,∞). Then

means that the values of f(x) become large (positive or negative) as x increases.

limx

f x

limx

f x

End BehaviorLet f be a function defined on some interval (-∞, a). Then

means that the values of f(x) become large (positive or negative) as x decreases.

limx

f x

limx

f x


Sketch a graph of a function with the following characteristics:

The function is continuous for all reals except 5.

lim 3x

f x

5

limx

f x

limx

f x

Calculating Limits at Infinity

Our book focuses on three ways:

1. Numerical Approach – Construct a table of values

2. Graphical Approach – Draw a graph

3. Analytic Approach – Use Algebra or calculus

First

Second

Example 1Use the graph and complete the table to find the limit (if it exists).

2

211

lim xxx

2

211

lim xxx

x 0 1 5 10 50 100 1000

f(x) -1 0 0.923 0.980 0.999998

0.99980.9992

As x increases, the value of the function approaches 1.

1


23 51lim x

xx

23 51lim x

xx

x -1000 -100 -50 -10 -5 -1 0

f(x) -3003 -303 -153 -32.778 -5UND-17.5

As x decreases, the value of the function decreases.


35limxx

35limxx

x 0 1 5 10 50 100 1000

f(x) UND 5 0.04 0.005 0.000000005

0.000005

0.00004

As x increases, the value of the function approaches 0.

0


10lim xx

10lim xx

x -9999 -5000 -1000 -100 -10 -1 0

f(x) 0.001 0.002 0.01 0.1 UND101

As x decreases, the value of the function approaches 0.

0

“Special Property” of Limits to Infinity

If A is any real number and r is a positive rational number then,

Furthermore, if r is such that xr is defined for x < 0, then

lim 0rAxx

lim 0rAxx


Use a table or graph to find the limit:

4

218 3 2

1lim x x

xx

Two Procedures for Analytically Determining Infinite Limits

If the function is a rational function or a radical/rational function:

1. Divide each term in the numerator and denominator by the highest power of x that occurs in the denominator.

2. Use basic limit laws and the “Special Property” of Infinite Limits to evaluate the limit.

OR

Use L’Hôpital’s Rule to evaluate the limit (Only if L’Hôpital’s Rule applies.)

Reminder

lim

limlim x

x

f xf x

g x g xx

lim

limlim x

x

f xf x

g x g xx

Example 1 (Procedure 1)

Analytically evaluate .

12 2

2 12

3 5 95 2 7

lim x

x

x xx xx

In order to use previous results, divide both the numerator and denominator by

the highest power of x appearing in

the fraction

23 5 92 2 2

25 2 72 2 2

limx x

x x x

x x

x x xx

5 92

722

3

5lim x x

x xx

Use “Direct Substitution” and previous results.

3 0 05 0 0

35

2

23 5 95 2 7

lim x xx xx

***Aside***

Analytically evaluate .

12 2

2 12

3 5 95 2 7

lim x

x

x xx xx

For this example, the limit’s value

does not change if x approaches

negative infinity.

23 5 92 2 2

25 2 72 2 2

limx x

x x x

x x

x x xx

5 92

722

3

5lim x x

x xx

3 0 05 0 0

35

2

23 5 95 2 7

lim x xx xx

L’Hôpital’s Rule applies since this is an

indeterminate form.


In order to use L’Hôpital’s Rule direct substitution must

result in 0/0 or ∞/∞.

2

23 5 95 2 7

lim x xx xx

Analytically evaluate 2

23 5 95 2 7

lim x xx xx

Differentiate the numerator and the denominator. 23 5 9d

dx x x 25 2 7ddx x x 6 5x 10 2x

Find the limit of the quotient of the derivatives.

6 510 2lim x

xx

This is still an indeterminate form, apply

L’Hôpital’s Rule again to the new limit.

Differentiate the new numerator and the denominator. 6 5d

dx x 10 2ddx x 6 10

Find the limit of the quotient of the

second derivatives.

610lim

x

610 3

5Since the result is finite or infinite, the

result is valid.

Example 2 (Procedure 1)Analytically evaluate .

3

595 57 30

1000lim x x

xx

13 5

5 15

95 57 301000

lim x

x

x xxx

In order to use

previous results, divide

both the numerator and denominator

by the highest power of x

appearing in the fraction

395 57 305 5 5

5 10005 5

limx x

x x x

x

x xx

95 57 302 4 5

100051

lim x x x

xx

Use “Direct Substitution” and previous results.

0 0 01 0

0

***Aside***Analytically evaluate .

3

595 57 30

1000lim x x

xx

13 5

5 15

95 57 301000

lim x

x

x xxx

395 57 30

5 5 5

5 10005 5

limx x

x x x

x

x xx

95 57 302 4 5

100051

lim x x x

xx

0 0 01 0

0



negative infinity.


3

595 57 30

1000lim x x

xx

L’Hôpital’s Rule applies since this is an indeterminate form.



3

595 57 30

1000lim x x

xx

Differentiate the numerator

and the denominator. 395 57 30d

dx x x

5 1000ddx x

2285 57x 45x

Find the limit of the quotient of the derivatives.2

4285 57

5lim x

xx

This is still an indeterminate form, apply

L’Hôpital’s Rule again to the new limit.

Differentiate the new numerator and the denominator. 2285 57d

dx x 45ddx x570x 320x


second derivatives.3

57020

lim xxx

This is still an indeterminate form, apply L’Hôpital’s Rule

again to the new limit.


22 13 5lim x

xx

12

12 13 5lim x

x

xxx

In order to use the previous result, divide

both the numerator and denominator by

the highest power of x

appearing in the fraction

12 2

12 13 5lim x

x

xxx

22 12 2

3 5limx

x xx

x xx

Use “Direct Substitution” and previous results

12

5

2

3lim x

xx

2 03 0

2Since as ,x x x

23

But, in order to simplify the

numerator, you must rewrite 1/x

***Aside***Analytically evaluate .

22 13 5lim x

xx

12

12 13 5lim x

x

xxx

1

2 2

12 13 5lim x

x

xxx

22 12 2

3 5limx

x xx

x xx


does change if x approaches

positive infinity.

12

5

2

3lim x

xx

2 03 0

2Since as ,x x x

23


Analytically evaluate .22 1

3 5lim xxx



22 13 5lim x

xx

Differentiate the numerator and the denominator.

22 1ddx x 3 5d

dx x 2

2

2 1

x

x 3

L’Hôpital’s Rule applies since this is an

indeterminate form.

Find the limit of the quotient of the derivatives.2

22 1

3limx

x

x

This is still an indeterminate form, apply L’Hôpital’s Rule

again to the new limit.Differentiate the new numerator

and the denominator. 2ddx x 23 2 1d

dx x 2 2

6

2 1

x

x


second derivatives.6

22 1

2lim x

xx

22 13lim x

xx

2

2

3 2 1lim x

xx

L’Hôpital’s Rule has failed to find a limit. This final result is almost

identical to the original. The first procedure is more applicable.

Example 4 (Procedure 1)Analytically evaluate the following limit:

1lim sin xxx

Now evaluate the limit: 1sin

1lim x

xx

Since the denominator is not a polynomial, we can not use the first procedure. We need to try

something new.

Rewrite the expression as a ratio in order to use the

first procedure.

1 11 sinx x

1sin

1x

x

Strategy: Rewrite one factor so its numerator

is 1.

Example 4 (Procedure 1)Analytically evaluate the following limit:

1lim sin xxx

In order to use L’Hôpital’s Rule direct substitution

must result in 0/0 or ∞/∞.

1sin

1lim x

xx 0

0

Differentiate the numerator and the

denominator.

1sinddx x

1ddx x

2 1cos xx2x

Find the limit of the quotient of the derivatives.

2 1

2

coslim xx

xx

1lim cos xx

L’Hôpital’s Rule applies since this is

an indeterminate form.

1

Since the result is finite or infinite,

the result is valid.

Rewrite the expression as a ratio in order to use

L’Hôpital’s Rule.

1 11 sinx x

1sin

1x

x

Strategy: Rewrite one factor so its numerator

is 1.

***Aside***Analytically evaluate the following limit:

1lim sin xxx

1sin

1lim x

xx 0

0

1sinddx x

1ddx x

2 1cos xx2x

2 1

2

coslim xx

xx

1lim cos xx 1

1 11 sinx x

1sin

1x

xFor this example, the limit’s value


negative infinity.

Day 45: November 10th

Objective: Determine (finite) limits at infinity, horizontal asymptotes of a graph if they exist, and infinite limits at infinity

• Homework Questions• Notes: Section 3.5• Conclusion

Homework: Read pgs. 198-204 and complete 3.5

http://www.haverford.k12.pa.us/site/handlers/filedownload.ashx?moduleinstanceid=8598&dataid=9613&FileName=Homework%20Problems.docx


Analytically evaluate each limit below:

25 3 21lim x x

xx

25 3 21lim x x

xx

Then y = 1 is a horizontal asymptote.

Horizontal Asymptotes and LimitsThe line y = L is called a horizontal asymptote of the

curve y = f(x) if L is finite and either

lim limx x

f x L or f x L

2

211

lim 1xxx

Since:

Procedure for Finding Horizontal Asymptotes

For a function f :

• Find the limit of the function as x goes to positive infinity.

• Find the limit of the function as x goes to negative infinity.

• If either of the above limits is finite, then they represent a horizontal asymptote(s) (remember to write the result as y = )

Examples Continued

For our previous examples:Function Horizontal Asymptotes

y = 3/5y = 0

y = 1NONE 25 3 2

1x x

xf x

2

23 5 95 2 7

x xx x

f x

3

595 57 30

1000x xx

f x

22 13 5

xxf x 2 2

3 3y and y

1sin xf x x

Whiteboard Challenge

On a calculator, graph

What is a characteristic of this graph that we have not

discussed?

2 23

x xxf x

Whiteboard Challenge

2 23

x xxf x

Slant/Oblique Asymptotes.

Oblique/Slant AsymptoteFor rational functions, slant asymptotes occur when the

degree of the numerator is one more than the degree of the denominator. In such a case the equation of the slant asymptote can be found by long division.

2 23

x xxf x

Degree = 2

Degree = 1

Procedure for Finding Oblique/Slant Asymptotes of a Rational Function

In a rational function f , if the degree of the numerator is one more than the degree of the denominator:

1. Perform Polynomial division.

2. Ignoring the remainder, the result is the oblique/slant asymptote.

(remember to write the result as y = )

ExampleAnalytically find the slant asymptote of 2 2

3x x

xf x

x

- 3

x

x2

-3x

2x

2

-6

4

RmPerform

Polynomial Division.

x2 – x – 2Thus: 2 2 4

3 32x xx xf x x

This means y = x + 2 is a slant asymptote because:

4 43 3lim 2 lim 2 lim lim 2 0 lim 2x xx x x x x

x x x x

Ignore the remainder

Asymptotes SummaryThe following asymptotes exists if…

Vertical: When there is a non-removable discontinuity (a value for x that makes the denominator 0 and the numerator non-zero)

Horizontal: When the limit as x approaches infinity (positive or negative), the value for y approaches a real number.

Slant: For a rational function, the degree of the numerator is one more than the degree of the denominator.

Section 3.5 – Limits at Infinity. Vertical Asymptotes and Limits When we investigated infinite limits and vertical asymptotes, we let x approach a number.

Documents

Section 3.5 – Limits at Infinity. Vertical Asymptotes and Limits When we investigated infinite limits and vertical asymptotes, we let x approach a number.