1.3 Evaluating Limits Analytically
After this lesson, you will be able to:
Evaluate a limit using the properties of limitsDevelop and use a strategy for finding limitsEvaluate a limit using dividing out and rationalizing techniquesEvaluate a limit using Squeeze Theorem
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.
Theorem 1.1 Some Basic Limits
Let b and c be real numbers and let n be a positive integer.
Examples 1:
limx c
b
b limx c
x
c lim n
x cx
nc
3lim 4x
42
limx
x
2 3
5limx
x
35 125Think of it graphically…( ) 4f x Let
As x approaches 3, f(x) approaches 4
3 x
Let ( )f x x
As x approaches 2, f(x) approaches 2
2 x
Let 3( )f x x(y scale was adjusted to fit)
As x approaches 5, f(x) approaches 125
5 x
Direct Substitution
0 0lim ( ) lim (0)
1 1x x
xf x fx
• Some limits can be evaluated by direct substituting 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 (or not approaches to zero) as the x approaches to a certain value.Example
11)(
xxxf
1 1
1lim ( ) lim (1)1 1 2 1x x
xf x fx
However
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:
and
Scalar multiple: lim [ ( )]x c
bf x
lim ( )x c
b f x
Sum or difference:
lim [ ( ) ( )]x c
f x g x
Lb
L K
Product:
lim [ ( ) ( )]x c
f x g x
LK
Quotient:
( )lim ( )x c
f xg x
L ,provided K 0K
Power:lim [ ( )]n
x cf x
Ln
lim ( ) Lx c
f x
lim ( ) Kx c
g x
Theorem 1.2 Properties of Limits
Note: The following assumptions are necessary to those properties of limits. You need provide all the counterexamples to some of the properties if the assumptions are not provided.
and
Scalar multiple: lim [ ( )]x c
bf x
lim ( )x c
b f x
Sum or difference:
lim [ ( ) ( )]x c
f x g x
Lb
L K
Product:
lim [ ( ) ( )]x c
f x g x
LK
Quotient:
( )lim ( )x c
f xg x
L ,provided K 0K
Power:lim [ ( )]n
x cf x
Ln
lim ( ) Lx c
f x
lim ( ) Kx c
g x
Limit of a Polynomial Function
Ans: The limit is 5
Example 2: 3 2
1lim 3 2 4
xx x
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.
If p is a polynomial function and c is a real number, then
If p(x) and q(x) are polynomial functions and r(x) = p(x)/q(x) and c is a real number such that q(c) ≠ 0, then
Theorem 1.3 Limits of Polynomial and Rational Function
)()(lim cpxpcx
)()()()(lim
cqcpcrxr
cx
Limit of a Rational Function
Example 3:
Ans: 1
Make sure the denominator doesn’t = 0 !
If the denominator had been 0, we would NOT have been able to use direct substitution.
224lim
2
1
xxx
x
Theorem 1.4 The Limit of a Function Involving a RadicalLet 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 n n
x cx c
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), then
x c
lim ( ) lim ( ) ( )x c x c
f g x f g x f L
Lx
Limit of a Composite Function-part aExample 4: Given
2( ) 2 3 1 and f x x x
3( ) 6 ,g x x
4lim ( )x
f x
a) Find
2
4lim 2 3 1x
x x
22(4) 3(4) 1
21
Direct substitution works here
Limit of a Composite Function -part b
21li ( )mx
g x
b) Find
Example 4: Given
2( ) 2 3 1 and f x x x
3( ) 6 ,g x x
3
21lim 6x
x
3 (21) 6
3 27 3
Direct substitution works here
Limit of a Composite Function -part c
4lim ( ( ))x
g f x
c) Find
Example 4: Given
2( ) 2 3 1 and f x x x
3( ) 6 ,g x x
4lim ( )x
g f x
(21)g
3
From part a, we know that the limit of f(x) as x approaches 4 is 21
Theorem 1.6 The Limits of Trig Functions
Let c be a real number in the domain of the given trigonometric function
cxcx
sinsinlim
1. cxcx
coscoslim
2.
cxcx
tantanlim
3. cxcx
cotcotlim
4.
cxcx
secseclim
5. cxcx
csccsclim
6.
Limits of Trig FunctionsExamples 5:
0lim tanx
x
lim ( cos )x
x x
2
0lim sinx
x
tan(0) 0
( )(cos ) ( )( 1)
2
0lim sinx
x
2sin 0 20 0
A Strategy for Finding Limits
Try Direct SubstitutionIf 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 or the denominator so that direct substitution works.Use a graph or table to reinforce your result.
Theorem 1.7 Functions That Agree at All But One Point
)(lim xfcx
Let c be a real number and let f (x) = g (x) for all x ≠ c in an open interval containing c. If exists, then
also exists and
)(lim xgcx
)(lim)(lim xgxfcxcx
Example- Factoring22
2lim 4x
xx
Example 6: 2
2lim 2 2x
xx x
Direct substitution at this point will give you 0 in the denominator. Using a bit of algebra, we can try to find the limit.
Factor
2
2lim x
x
2 2x x
1
2
1lim 2x x
Now direct substitution will work
14
Graph on your calculator and use the table to check your result
Example- Factoring2
24
5 4lim 2 8x
x xx x
Example 7: 4
( 4)lim x
x
( 1)( 4)
xx
( 2)x
Direct substitution results in 0 in the denominator. Try factoring.
4
( 1)lim ( 2)x
xx
Now direct substitution will work.
136 2
Use your calculator to reinforce your result.
Example3
1
1lim1x
xx
Example 8:
Direct substitution results in 0 in the denominator. Try factoring.
2
1
1 1lim
1x
x x x
x
Sum of cubes Not
factorable
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.Observe how the right limit goes to off to positive infinity
and the left limit goes to negative infinity.
Example- Rationalizing Technique
Example 9:
0
1 1limx
xx
Direct substitution results in 0 in the denominator. I see a radical in the numerator. Let’s try rationalizing the numerator.
0
1 1 1 1lim
1 1x
x x
x x
Multiply the top and bottom by the conjugate of the numerator.
0
1 1lim
1 1x
x
x x
Note: It was convenient NOT to distribute on the bottom, but you did need to FOIL on the top
0limx
x
x 1 1x
0
12
1lim1 1x x
Now direct substitution will work
Go ahead and graph to verify.
Theorem 1.8 The Squeeze Theorem
Lxfcx
)(lim
If h(x) ≤ f (x) ≤ g(x) for all x in an open interval containing c, except possibly at c itself, and if
)(lim)(lim xgLxhcxcx
then
Two Special Trigonometric Limits
In your text, read about the Squeeze Theorem on page 65. Following the
Squeeze Theorem are the proofs of two special trig limits…I will not expect you to
be able to prove the two limits, so you’ll just want to memorize them. The next slide will give them to you and then we’ll use them in
a few examples.
Trig Limits
0
sinlim 1x
xx
(A star will indicate the need to memorize!!!)
0
1 coslim 0x
xx
Think of this as the limit as “something” approaches 0 of the sine of “something” over the same “something” is equal to 1.
0
sin lim 1
Example- Using Trig Limits
0
sin 5limx
xx
Example 11:
Before you decide to even use a special trig limit, make sure that direct substitution won’t work. In this case, direct substitution really won’t work, so let’s try to get this to look like one of those special trig limits.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.
0
sin 5lim 55x
xx
This 5 is a constant and can be pulled out in front of the limit.
0
sin 5li5 m5x
xx
5 15
=1
Example 10, page 66, in your text is another example.
Example
0
sin 3lim2x
xx
Example 12
Direct substitution won’t work. We can use the sine trig limit, but first we’ll have to use some algebra since we need the bottom to be a 3x.To create a 3x on the bottom, we’ll multiply the bottom by 3/2. To be “fair”, we’ll have to multiply the top by 3/2 as well. Watch how I would do it.
0
32
32
sin 3lim2x
xx
0
3 sin 3lim2 3x
xx
= 1
232
31
Example
2
sinlimx
xx
Example 13
This example was thrown in to keep you on your toes. Direct substitution works at this point since the bottom of the fraction will not be 0 when you use π/2.
2
2
sin
Looking at the unit circle, this value is 1
2
1 2
Example
0
1 coslimx
xx
Example
14
This is the 2nd special trig limit and you should know that this limit is 0. Let’s prove it by using the 1st trig limit.
0
1 cos 1 coslim1 cosx
x xx x
2
0
1 coslim(1 cos )x
xx x
Again, in this case, it’s best not to distribute on the bottom. You’ll see why it helps to leave it in factored form for now.
2
0
sinlim(1 cos )x
xx x
Use the Pythagorean Trig Identity
0
sin sinlim(1 cos )x
x xx x
Be
creativeSpecial Trig Limit
sin 0 01 11 cos 0 2
0