finding limits algebraically in section 12-1, we used calculators and graphs to guess the values of...

Finding Limits Algebraically

In Section 12-1, we used calculators and graphs to guess the values of limits.

• However, we saw that such methods don’t always lead to the correct answer.

In this section, we use algebraic methods to find limits exactly.

Limit Laws

Limit Laws

We use the following properties of limits, called the Limit Laws, to calculate limits.

• Suppose that c is a constant and that the following limits exist:

lim ( ) and lim ( )x a x a

f x g x

Limit Laws 1, 2, and 3

Then,

1 Limit of a Sum

2Limit of

a Difference

3Limit of

a Constant Multiple

lim ( ) ( )

lim ( ) lim ( )x a

x a x a

f x g x

f x g x

lim ( ) ( )

lim ( ) lim ( )x a

x a x a

f x g x

f x g x

lim ( ) lim ( )x a x a

cf x c f x

Limit Laws 4 and 5

4Limit of

a Product

5Limit of

a Quotient

lim ( ) ( )

lim ( ) lim ( )x a

x a x a

f x g x

f x g x

lim ( )( )lim( ) lim ( )

if lim ( ) 0

x a

x ax a

x a

f xf xg x g x

g x

Limit Laws

These five laws can be stated verbally as follows:

1. The limit of a sum is the sum of the limits.

2. The limit of a difference is the difference of the limits.

Limit Laws

3. The limit of a constant times a function is the constant times the limit of the function.

4. The limit of a product is the product of the limits.

5. The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0).

Limit Laws

It’s easy to believe that these properties are true.

• For instance, if f(x) is close to L and g(x) is close to M, it is reasonable to conclude that f(x) + g(x) is close to L + M.

• This gives us an intuitive basis for believing that Law 1 is true.

Limit Laws

If we use Law 4 (Limit of a Product) repeatedly with g(x) = f(x) , we obtain the following Law 6 for the limit of a power.

A similar law holds for roots.

Limit Laws 6 and 7

6where n is a positive integer

Limit of a Power

7where n is a positive integer

Limit of a Root

lim ( ) lim ( )nn

x a x af x f x

lim ( ) lim ( )n n

x a x af x f x

• If n is even, we assume that .

lim 0x a

f x

Limit Laws

In words, these laws say:

6. The limit of a power is the power of the limit.

7. The limit of a root is the root of the limit.

E.g. 1—Using the Limit Laws

Use the Limit Laws and the graphs of f and g in to evaluate these limits, if they exist.

2

1

2

3

1

(a) lim ( ) 5 ( )

(b) lim ( ) ( )

( )(c) lim( )

(d) lim ( )

x

x

x

x

f x g x

f x g x

f xg x

f x

E.g. 1—Using the Limit Laws Example (a)

We see that:

• Therefore,

2 2lim ( ) 1 and lim ( ) 1x x

f x g x

2

2 2

2 2

lim ( ) 5 ( )

lim ( ) lim 5 ( )

lim ( ) 5 lim ( )

1 5( 1) 4

x

x x

x x

f x g x

f x g x

f x g x

E.g. 1—Using the Limit Laws Example (b)

We see that .

However, does not exist because the left- and right-hand limits are different:

1lim ( ) 2x

f x

1lim ( )x

g x

1

1

lim ( ) 2

lim ( ) 1

x

x

g x

g x

E.g. 1—Using the Limit Laws Example (b)

So, we can’t use Law 4. • The given limit does not exist—since

the left-hand limit is not equal to the right-hand limit.

E.g. 1—Using the Limit Laws Example (c)

The graphs show that:

• As the limit of the denominator is 0, we can’t use Law 5.

• The given limit does not exist—as the denominator approaches 0 while the numerator approaches a nonzero number.

2

2

lim ( ) 1.4

andlim ( ) 0

x

x

f x

g x

E.g. 1—Using the Limit Laws Example (d)

Since , we use Law 6 to get:

1lim ( ) 2x

f x

3

1

3

1

3

lim ( )

lim ( )

2 8

x

x

f x

f x

Applying the Limit Laws

Some Special Units

In applying the Limit Laws, we need to use four special limits.

1

2

3

4

lim

lim

lim ( is a positive integer)

lim ( is a positive integer and 0)

x a

x a

n n

x a

n n

x a

c c

x a

x a n

x a n a

Some Special Units

Special Limits 1 and 2 are intuitively obvious.

• Looking at the graphs of y = c and y = x will convince you of their validity.

Special Limits 3 and 4 are special cases of Limit Laws 6 and 7.


Evaluate the following limits and justify each step.

2

5

3 2

2

(a) lim(2 3 4)

2 1(b) lim5 3

x

x

x x

x xx

E.g. 2—Using the Limit Laws Example (a)

2

5

2

5 5 5

2

5 5 5

2

(Difference & Sum)

(Constant Multiple)

(SpecialLimits 3,2,&1)

lim(2 3 4)

lim(2 ) lim(3 ) lim 4

2lim 3lim lim 4

2(5 ) 3(5)

3

4

9

x

x x x

x x x

x x

x x

x x


We start by using Law 5.

However, its use is fully justified only at the final stage—when we see that:

• The limits of the numerator and denominator exist.

• The limit of the denominator is not 0.

Example (b)


3 2

2

3 2

2

2

3 2

2 2 2

2 2

(Sums,Differences,

& Constant Multiples)

2 1lim5 3

lim ( 2 1)

lim (5 3 )

lim 2 lim lim 1

lim 5 3 lim

x

x

x

x x x

x x

x xx

x x

x

x x

x

Example (b)


3 2

(Special Limits 3, 2, & 1)( 2) 2( 2) 1

5 3( 2)

111

Example (b)

Direct Substitution

If we let f(x) = 2x2 – 3x + 4, then f(5) = 39.

• In Example 2 (a), we found that .

• That is, we would have gotten the correct answer by substituting 5 for x.

• Similarly, direct substitution provides the correct answer in part (b).

5lim ( ) 39x

f x

Direct Substitution

• The functions in Example 2 are a polynomial and a rational function, respectively.

• Similar use of the Limit Laws proves that direct substitution always works for such functions.

• We state this fact as follows.

Limits by Direct Substitution

If f is a polynomial or a rational function and a is in the domain of f, then

lim ( ) ( )x a

f x f a

Continuous Functions

Functions with this direct substitution property are called continuous at a.

E.g. 3—Finding Limits by Direct Substitution

Evaluate the following limits.

3

3

2

41

(a) lim(2 10 8)

5(b) lim2

x

x

x x

x xx

E.g. 3—Direct Substitution

The function f(x) = 2x3 – 10x – 8 is a polynomial.

• So, we can find the limit by direct substitution:

3

3lim(2 10 8) 2(3) 10(3) 8

16x

x x

Example (a)

E.g. 3—Direct Substitution

The function f(x) = (x2 + 5x)/(x4 + 2) is a rational function.

Also, x = –1 is in its domain (because the denominator is not zero for x = –1. )

• So, we can find the limit by direct substitution:2 2

4 41

5 ( 1) 5( 1) 4lim2 ( 1) 2 3x

x xx

Example (b)

Finding Limits Using Algebra and the Limit Laws

Finding Limits Using Algebra and the Limit Laws

As we saw in Example 3, evaluating limits by direct substitution is easy.

However, not all limits can be evaluated this way.

• Most situations in which limits are useful require us to work harder to evaluate the limit.

• The next three examples illustrate how we can use algebra to find limits.

E.g. 4—Finding a Limit by Canceling a Common Factor

Find

• Let f(x) = (x – 1)/(x2 – 1). • We can’t find the limit by substituting x = 1

because f(1)isn’t defined. • Nor can we apply Law 5 since the limit of

the denominator is 0. • Instead, we need to do some preliminary algebra.

21

1lim1x

xx


We factor the denominator as a difference of squares:

• The numerator and denominator have a common factor of x – 1.

• When we take the limit as x approaches 1, we have x ≠ 1, and so x – 1 ≠ 0.

2

1 11 ( 1)( 1)

x xx x x


So, we can cancel the common factor and compute the limit as follows:

• This calculation confirms algebraically the answer we got numerically and graphically in Example 1 in Section 12-1.

21 1

1

1 1lim lim1 ( 1)( 1)

1 1 1lim1 1 1 2

x x

x

x xx x x

x

E.g. 5—Finding a Limit by Simplifying

Evaluate

• We can’t use direct substitution to evaluate this limit—as the limit of the denominator is 0.

• So, we first simplify the limit algebraically.

2

0

(3 ) 9limh

hh

E.g. 5—Finding a Limit by Simplifying

2 2

0 0

2

0

0

(3 ) 9 (9 6 ) 9lim lim

6lim

lim(6 )

6

h h

h

h

h h hh h

h hhh

E.g. 6—Finding a Limit by Rationalizing

Find

• We can’t apply Law 5 immediately—as the limit of the denominator is 0.

• Here, the preliminary algebra consists of rationalizing the numerator.

2

20

9 3limt

tt


2

20

2 2

2 20

2

2 20

2

2 20

9 3lim

9 3 9 3lim9 3

( 9) 9lim( 9 3)

lim( 9 3)

t

t

t

t

tt

t tt t

tt t

tt t


• This calculation confirms the guess we made in Example 2 in Section 12-1.

20

2

0

1lim9 3

1

lim( 9) 3

1 13 3 6

t

t

t

t

Using Left- and Right-Hand Limits


Some limits are best calculated by first finding the left- and right-hand limits.

• The following theorem is a reminder of what we discovered in Section 12-1.

• It says that a two-sided limit exists if and only if both of the one-sided limits exist and are equal.


• When computing one-sided limits, we use the fact that the Limit Laws also hold for one-sided limits.

lim ( )

if and only iflim ( ) lim ( )

x a

x a x a

f x L

f x L f x

E.g. 7—Comparing Right and Left Limits

Show that .

• Recall that:

0lim | | 0x

x

if 0if 0

x xx

x x


Since | x | = x for x > 0, we have:

For x < 0, we have | x | = –x, and so

• Therefore,

0 0lim lim 0x x

x x

0 0lim lim ( ) 0x x

x x

0lim 0x

x


The result of Example 7 looks plausible from this figure.


Prove that does not exist.

• Since | x | = x for x > 0 and | x | = –x for x < 0, we have:

• Since the right-hand and left-hand limits exist and are different, it follows that the limit does not exist.

0limx

xx

0 0 0

0 0 0

lim lim lim 1 1

lim lim lim ( 1) 1

x x x

x x x

x xx xx xx x


The graph of the function is shown here and supports the limits we found.

E.g. 9—Limit of a Piecewise-Defined Function

Let

Determine whether exists.

4 if 4( )8 2 if 4

x xf xx x

4lim ( )x

f x


Since for x > 4, we have:

Since f(x) = 8 – 2x for x < 4, we have:

• The right- and left-hand limits are equal.• Thus, the limit exists and .

( ) 4f x x

4 4lim ( ) lim 4 4 4 0x x

f x x

4 4lim ( ) lim (8 2 ) 8 2 4 0x x

f x x

4lim ( ) 0x

f x


The graph of f is shown.

finding limits algebraically in section 12-1, we used calculators and graphs to guess the values of...

Documents