Transcript Evaluating Limits Analytically

Evaluating Limits Analytically
Section 1.3
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After this lesson, you will be able to:
evaluate a limit using the properties of limits
develop and use a strategy for finding limits
evaluate a limit using dividing out and
rationalizing techniques
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Limits Analytically
In the previous lesson, you learned how to find limits
numerically and graphically. In this lesson you will be shown
how to find them analytically…using algebra or calculus.
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Theorem 1.1 Some Basic Limits
Let b and c be real numbers and let n be a positive integer.
lim b  b
xc
Examples
lim 4 
x 3
____
Let f ( x)  4
Think of it
graphically
3 x
As x approaches 3,
f(x) approaches 4
lim x  c
lim x  c
lim x  ____
x2
lim x  ____
x 5
Let f ( x)  x
Let f ( x)  x3
xc
2x
As x approaches 2,
f(x) approaches 2
n
n
x c
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y scale was
adjusted to fit
5 x
As x approaches 5,
f(x) approaches 125
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Direct Substitution
•Some limits can be evaluated by direct substitution for x.
•Direct substitution works on continuous functions.
•Continuous functions do NOT have any holes, breaks or gaps.
Note: Direct substitution is valid for all polynomial functions
and rational functions whose denominators are not zero.
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Theorem 1.2 Properties of Limits
Let b and c be real numbers, let n be a positive integer, and let f and g
be functions with the following limits:
lim f ( x)  L
x c
Scalar multiple:
Sum or difference:
Product:
Quotient:
Power:
and
lim g ( x)  K
x c
lim [bf ( x)]  b lim f ( x)  bL
x c
x c
lim [ f ( x)  g ( x)]  L  K
x c
lim [ f ( x) g ( x)]  LK
x c
lim
x c
L
f ( x)
, provided K  0

K
g ( x)
lim [ f ( x)]n  Ln
x c
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Limit of a Polynomial Function
Example: lim  3x3  2 x 2  4  
x 1
Since a polynomial function is a continuous function, then we know the limit
from the right and left of any number will be the same. Thus, we may use
direct substitution.
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Limit of a Rational Function
Make sure the denominator doesn’t = 0 !
2x  3
Example: lim
x 3 x  5
If the denominator had been 0, we would not have been able to use direct substitution.
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Theorem 1.4 The Limit of a Function Involving a
Radical
Let n be a positive integer. The following limit is valid for
all c if n is odd, and is valid for c > 0 if n is even.
lim x  c
n
n
x c
So we can use direct substitution again, as long as c is in the
domain of the radical function.
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Theorem 1.5 The Limit of a Composite Function
If f and g are functions such that lim g(x) = L and lim f(x) = f(L),
x c
xL
then


lim f  g ( x)   f lim g ( x)  f ( L)
xc
xc
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Limit of a Composite Function-part a
Example: Given
find
f ( x)  x  6
3
and
g ( x)  2x  3x  1,
2
lim f  g  x  
x 4
a) First find
lim g ( x)
x4


 lim 2 x 2  3x  1
x 4

Direct substitution works here
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Limit of a Composite Function -part b
b) Then find
lim f ( x )
x  21
 lim 3 x  6
x 21
Direct substitution works here, too.

*********************************************
Therefore,

lim f  g  x    f lim g ( x)
x 4
 f(

x4

)
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Limits of Trig Functions
If c is in the domain of the given trig function, then
lim sin x  sin c
lim csc x  csc c
lim cos x  cos c
lim sec x  sec c
lim tan x  tan c
lim cot x  cot c
x c
xc
x c
x c
x c
x c
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Limits of Trig Functions
Examples:
lim tan x 
x0
lim ( x cos x) 
x 
lim sin 2 x 
x 0
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Limits of Trig Functions
Examples:
lim
3
x
2
 sin x  cos x  
lim tan 2 x 
x

4
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Finding Limits
Try Direct Substitution
If the limit of f(x) as x approaches c cannot be
evaluated by direct substitution, try to divide out
common factors or to rationalize the numerator
so that direct substitution works.
Use a graph or table to reinforce your result.
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Example 1- Factoring
Factor the denominator
2 x
 lim
Example* : lim
2
x 2
x2 x  4
2 x
 x  2  x  2 
1
Direct substitution at this point
will give you 0 in the denominator:
22
0

22  4 0
Now direct substitution will work
 lim
x2
2 x
 x  2  x  2
1
 lim
x 2
 x  2
1

4
Graph on your calculator
and use the table to check
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your result
Example 2- Factoring
2
x
 5x  4
Example: lim

2
x4 x  2 x  8
Direct substitution results in the
indeterminate form 0/0.
Try factoring.
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Example: Limit DNE
Sum of cubes
Example*:
x 1
lim
 lim
x 1 x  1
x1
3
Not factorable
 x  1  x2  x  1
x 1
Direct substitution results in 0 in
the denominator. Try factoring.
None of the factors can be divided out,
so direct substitution still won’t work.
The limit DNE. Verify the
result on your calculator.
The limits from the right and left do not equal each
other, thus the limit DNE.
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Observe how the right limit goes to off to positive infinity and the left limit goes to negative infinity.
Example 1- Rationalizing Technique
Example*:
lim
x 0
x 1 1
x
First, we will try direct substitution:
lim
x 0
x  1 1

x
0  1 1 1 1 0


0
0
0
Indeterminate
Form
Plan: Rationalize the numerator to come up with a related
function that is defined at x = 0.
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Example 1- Rationalizing Technique
Example*:
lim
x 0
x 1 1

x

lim
x 0
Multiply the numerator
and denominator by the
conjugate of the
numerator.
Note: It was convenient NOT to
distribute in the denominator, but
you did need to FOIL in the
numerator.



x  1  1
x 1 1
x
 x  1  1
 lim
x 0
x
 lim
x 0
 lim
x 0
x 1 1
x



x 1 1
x

x 1 1
1
1

x 1 1 2
Now direct substitution will work
Go ahead and graph to verify.
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Example 2- Rationalizing Technique
Example:
lim
x 4
x 5 3
x4
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Two Very Important Trig Limits
sin x
lim
1
x 0
x
lim
sin
0
1
(A star will indicate the need to memorize!!!)
1  cos x
lim
0
x 0
x
lim
2 x 0
1  cos  2 x 
2x
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0
Example 1- Using Trig Limits
Example*:
sin x
lim
1
x 0
x
sin 5 x lim 5  sin 5 x  5 lim sin 5 x
lim
 x 0
x 0
5x
5
x
x 0
x
Before you decide to even use
a special trig limit, make sure
that direct substitution won’t
work. In this case, direct
substitution won’t work, so
let’s try to get this to look like
one of those special trig
limits.
This 5 is a constant
and can be pulled out
in front of the limit.
Note: x  0 iff 5 x  0
sin 5 x
5 lim
5 x 0
5x
 5 1
Equals 1
5
Now, the 5x is like the heart. You will need the bottom to also be
5x in order to use the trig limit. So, multiply the top and bottom by
5. You won’t have changed the fraction. Watch how to do it.
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Example 2
sin 3 x
lim

x 0
2x
Try to get into the form:
lim
x 0
sin x
1
x
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Example 3
sin x
lim

x  2
x
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Example 4
3  3cos x
lim

x 0
x
Get into the form:
1  cos x
0
x 0
x
lim
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Example 5
tan x
lim

x 0
x
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Homework
Section 1.3: page 67 #1, 5-39 odd, 49-61 odd, 67-77 odd
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