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