1 evaluating limits analytically section 1.3. 2 after this lesson, you will be able to: evaluate a...

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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 limitsdevelop and use a strategy for finding limitsevaluate 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.

limx c

b b

limx c

x c

lim n n

x cx c

Think 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

3lim 4x

____ 2

limx

x

____3

5limx

x

____Examples

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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:

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 x

g x

L,provided K 0

K

Power:lim [ ( )]n

x cf x

Ln

and

lim ( ) Lx c

f x

lim ( ) Kx c

g x

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Limit of a Polynomial Function

Example: 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.

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Limit of a Rational Function

Example:

3

2 3lim

5x

x

x

Make sure the denominator doesn’t = 0 !

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 n n

x cx 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), then

x c x L

lim ( ) lim ( ) ( )x c x c

f g x f g x f L

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Limit of a Composite Function-part a

Example: Given and ,

2( ) 2 3 1g x x x 3( ) 6f x x

a) First find 4

lim ( )x

g x

2

4lim 2 3 1x

x x

Direct substitution works here

find 4

limx

f g x

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Limit of a Composite Function -part b

21li ( )mx

f x

b) Then find

3

21lim 6x

x

Direct substitution works here, too.

*********************************************Therefore,

4limx

f g x

4lim ( )x

f g x

( )f

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Limits of Trig Functions

If c is in the domain of the given trig function, then

limsin sinx c

x c

lim cos cosx c

x c

lim tan tanx c

x c

lim csc cscx c

x c

limsec secx c

x c

lim cot cotx c

x c

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Limits of Trig Functions

Examples:

0lim tanx

x

lim ( cos )x

x x

2

0lim sinx

x

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Limits of Trig Functions

Examples:

3

2

lim sin cosx

x x

2

4

lim tanx

x

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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

22

2lim

4x

x

x

Example* : 2

2lim

2 2x

x

x x

Factor the denominator

2

2lim x

x

1

22 xx

2

1lim

2x x

Now direct substitution will work 1

4 Graph on your

calculator and use the table to check your result

Direct substitution at this point will give you 0 in the denominator:

2

02

2

2

4 0

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Example 2- Factoring

2

24

5 4lim

2 8x

x x

x x

Example:

Direct substitution results in the indeterminate form 0/0. Try factoring.

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Example: Limit DNE

3

1

1lim

1x

x

x

Example*:

Direct substitution results in 0 in the denominator. Try factoring.

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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.

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Example 1- Rationalizing TechniqueExample*:

0

1 1limx

x

x

First, we will try direct substitution:

0

01 1 1 1 1 1 0

0 0i

0l mx

x

x

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*:

0

1 1limx

x

x

0

1 1 1 1lim

1 1x

x x

x x

Multiply the numerator and denominator by the conjugate of the numerator.

0

1 1lim

1 1x

x

x x

Note: It was convenient NOT to distribute in the denominator, but you did need to FOIL in the numerator.

0limx

x

x 1 1x

0

1

2

1lim

1 1x x

Now direct substitution will work Go ahead and graph to

verify.

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Example 2- Rationalizing TechniqueExample:

4

5 3lim

4x

x

x

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Two Very Important Trig Limits

0

sinlim 1x

x

x

(A star will indicate the need to memorize!!!)

0

1 coslim 0x

x

x

0

sin lim 1

2 0

1 cos 2lim 0

2x

x

x

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Example 1- Using Trig Limits

0

sin 5limx

x

x

Example*:

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.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

5

5x

x

x

5 1

5

0

sin 5li5 m

5x

x

x

0

sinlim 1x

x

x

Note: 0 5 0x iff x

5 0

sin 5lim5

5x

x

x

Equals 1

This 5 is a constant and can be pulled out in front of the limit.

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Example 2

0

sin 3lim

2x

x

x

Try to get into the form: 0

sinlim 1x

x

x

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Example 3

2

sinlimx

x

x

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Example 4

0

3 3coslimx

x

x

0

1 coslim 0x

x

x

Get into the form:

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Example 5

0

tanlimx

x

x

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Homework

Section 1.3: page 67 #1, 5-39 odd, 49-61 odd, 67-77 odd