Upload
dana-ringer
View
222
Download
0
Embed Size (px)
Citation preview
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
White Board Challenge
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
Example 2Use the graph and complete the table to find the limit (if it exists).
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.
Example 3Use the graph and complete the table to find the limit (if it exists).
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
Example 4Use the graph and complete the table to find the limit (if it exists).
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
White Board Challenge
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.
Example 1 (Procedure 2)
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
For this example, the limit’s value
does not change if x approaches
negative infinity.
Example 2 (Procedure 2)Analytically evaluate .
3
595 57 30
1000lim x x
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 ∞/∞.
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
Find the limit of the quotient of the
second derivatives.3
57020
lim xxx
This is still an indeterminate form, apply L’Hôpital’s Rule
again to the new limit.
Example 3 (Procedure 1)Analytically evaluate .
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
For this example, the limit’s value
does change if x approaches
positive infinity.
12
5
2
3lim x
xx
2 03 0
2Since as ,x x x
23
Example 3 (Procedure 2)
Analytically evaluate .22 1
3 5lim xxx
In order to use L’Hôpital’s Rule direct substitution must
result in 0/0 or ∞/∞.
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
Find the limit of the quotient of the
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
does not change if x approaches
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
White Board Challenge
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.