LIMITS AND DERIVATIVESLIMITS AND DERIVATIVES

2

We have used calculators

and graphs to guess the values

of limits. However, we have learned that such methods

don’t always lead to the correct answer.

LIMITS AND DERIVATIVES

2.3Calculating Limits

Using the Limit Laws

LIMITS AND DERIVATIVES

In this section, we will:

Use the Limit Laws to calculate limits.

Suppose that c is a constant and

the limits and

exist.

lim ( )x a

f x

lim ( )x a

g x

THE LIMIT LAWS

Then,

4.lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x

1.lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x

THE LIMIT LAWS

2.lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x

3.lim ( ) lim ( )x a x a

cf x c f x

lim ( )( )5.lim lim ( ) 0

( ) lim ( )x a

x a x ax a

f xf xif g x

g x g x

These laws can be

stated verbally.

THE LIMIT LAWS

The limit of a sum is the sum

of the limits.

THE SUM LAW

lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x

The limit of a difference is the

difference of the limits.

THE DIFFERENCE LAW

lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x

The limit of a constant times a

function is the constant times the

limit of the function.

THE CONSTANT MULTIPLE LAW

lim ( ) lim ( )x a x a

cf x c f x

The limit of a product is the

product of the limits.

THE PRODUCT LAW

lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x

The limit of a quotient is the quotient

of the limits (provided that the limit of

the denominator is not 0).

THE QUOTIENT LAW

lim ( )( )lim lim ( ) 0

( ) lim ( )x a

x a x ax a

f xf xif g x

g x g x

It is 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 the Sum Law is true.

THE LIMIT LAWS

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

a.

b.

c.

2

lim ( ) 5 ( )x

f x g x

USING THE LIMIT LAWS Example 1

1

lim ( ) ( )x

f x g x

2

( )lim

( )x

f x

g x

From the graphs, we see that

and .

Therefore, we have:

2lim ( ) 1x

f 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

USING THE LIMIT LAWS Example 1 a

2lim ( ) 1x

g x

We see that .

However, does not exist—because

the left and right limits are different: and

So, we can’t use the Product Law for the desired limit.

1lim ( ) 2x

f x

1lim ( )x

g x

1lim ( ) 2x

g x

USING THE LIMIT LAWS Example 1 b

1lim ( ) 1x

g x

However, we can use the Product Law

for the one-sided limits:

and

The left and right limits aren’t equal.

So, does not exist.

1

lim ( ) ( )x

f x g x

USING THE LIMIT LAWS Example 1 b

1lim[ ( ) ( )] 2 ( 2) 4x

f x g x

1lim[ ( ) ( )] 2 ( 1) 2x

f x g x

The graphs show that and

.

As the limit of the denominator is 0, we

can’t use the Quotient Law. does not exist. This is because the

denominator approaches 0 while the numerator approaches a nonzero number.

2lim ( ) 1.4x

f x

USING THE LIMIT LAWS Example 1 c

2

( )lim

( )x

f x

g x

2lim ( ) 0x

g x

If we use the Product Law repeatedly

with f(x) = g(x), we obtain the Power

Law.

where n is a positive integer

6.lim ( ) lim ( )nn

x a x af x f x

THE POWER LAW

In applying these six limit laws, we

need to use two special limits.

These limits are obvious from an intuitive point of view. State them in words or draw graphs of y = c and y = x.

7.lim

x ac c

USING THE LIMIT LAWS

8.limx a

x a

If we now put f(x) = x in the Power Law

and use Law 8, we get another useful

special limit.

where n is a positive integer.

9.lim n n

x ax a

USING THE LIMIT LAWS

A similar limit holds for roots.

If n is even, we assume that a > 0.

10.lim n n

x ax a

USING THE LIMIT LAWS

More generally, we have the

Root Law.

where n is a positive integer.

If n is even, we assume that .

11.lim ( ) lim ( )n nx a x a

f x f x

lim ( ) 0x a

f x

THE ROOT LAW

Evaluate the following limits and

justify each step.

a.

b.

2

5lim(2 3 4)x

x x

USING THE LIMIT LAWS Example 2

3 2

2

2 1lim

5 3x

x x

x

(by Laws 2 and 1)

2

5

2

5 5 5

2

5 5 5

2

lim(2 3 4)

lim(2 ) lim3 lim 4

2lim 3lim lim 4

2(5 ) 3(5) 4

39

x

x x x

x x x

x x

x x

x x

USING THE LIMIT LAWS Example 2 a

(by Law 3)

(by Laws 9, 8, and 7)

We start by using the Quotient Law.

However, its use is fully justified only

at the final stage. That is when we see that the limits of the numerator

and denominator exist and the limit of the denominator is not 0.

USING THE LIMIT LAWS Example 2 b

3 2

2

3 2

2

2

3 2

2 2 2

2 2

3 2

2 1lim

5 3

lim( 2 1)

lim(5 3 )

lim 2 lim lim 1

lim 5 3 lim

( 2) 2( 2) 1 1

5 3( 2) 11

x

x

x

x x x

x x

x x

x

x x

x

x x

x

USING THE LIMIT LAWS Example 2 b

(by Law 5)

(by Laws 1, 2, and 3)

(by Laws 9, 8, and 7)

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

In other words, we would have gotten the correct answer in Example 2 a by substituting 5 for x.

Similarly, direct substitution provides the correct answer in Example 2 b.

USING THE LIMIT LAWS Note

The functions in the example are

a polynomial and a rational function,

respectively. Similar use of the Limit Laws proves that direct

substitution always works for such functions.

USING THE LIMIT LAWS Note

We state this fact as follows.

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

DIRECT SUBSTITUTION PROPERTY

Functions with the Direct Substitution Property

are called ‘continuous at a.’

However, not all limits can be evaluated by

direct substitution—as the following examples

show.

DIRECT SUBSTITUTION PROPERTY

Find

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

f(1) isn’t defined. We can’t apply the Quotient Law because the limit

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

2

1

1lim .

1x

x

x

USING THE LIMIT LAWS Example 3

We factor the numerator 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 and so .

2 1 ( 1)( 1)

1 ( 1)

x x x

x x

USING THE LIMIT LAWS Example 3

1x 1 0x

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

2

1

1

1

1lim

1( 1)( 1)

lim( 1)

lim( 1)

1 1

2

x

x

x

x

xx x

x

x

USING THE LIMIT LAWS Example 3

The limit in the example arose in Section 2.1

when we were trying to find the tangent to the

parabola y = x2 at the point (1, 1).

USING THE LIMIT LAWS Example 3

In the example, we were able to compute

the limit by replacing the given function

f(x) = (x2 - 1)/(x - 1) by a simpler function

with the same limit, g(x) = x + 1. This is valid because f(x) = g(x) except when x = 1

and, in computing a limit as x approaches 1, we don’t consider what happens when x is actually equal to 1.

USING THE LIMIT LAWS Note

In general, we have the following

useful fact.

If f(x) = g(x) when , then , provided the limits

exist.

x alim ( ) lim ( )x a x a

f x g x

USING THE LIMIT LAWS Note

Find where .

Here, g is defined at x = 1 and . However, the value of a limit as x approaches 1 does

not depend on the value of the function at 1. Since g(x) = x + 1 for , we have

.

1lim ( )x

g x

1 1( )

1

x if xg x

if x

USING THE LIMIT LAWS Example 4

(1)g

1x

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

g x x

Note that the values of the functions in

Examples 3 and 4 are identical except when

x = 1.

So, they have the same limit as x

approaches 1.

USING THE LIMIT LAWS

Evaluate

If we define , then, we can’t compute by letting h = 0 since F(0) is undefined.

However, if we simplify F(h) algebraically, we find that:

2

0

(3 ) 9lim .h

h

h

USING THE LIMIT LAWS Example 5

2(3 ) 9( )

hF h

h

0lim ( )h

F h

2

2

( )

(9 6 ) 9

6

6

F h

h h

h

h h

hh

Recall that we consider only when letting h approach 0.

Thus,

0h

2

0

0

(3 ) 9lim

lim(6 )

6

h

h

h

hh

USING THE LIMIT LAWS Example 5

Find

We can’t apply the Quotient Law immediately—since the limit of the denominator is 0.

Here, the preliminary algebra consists of rationalizing the numerator.

2

20

9 3lim .t

t

t

USING THE LIMIT LAWS Example 6

2

20

2 2

2 20

2

2 20

2

2 20

20

2

0

9 3lim

9 3 9 3lim

9 3

( 9) 9lim

( 9 3)

lim( 9 3)

1lim

9 31 1 1

3 3 6lim( 9) 3

t

t

t

t

t

t

t

t

t t

t t

t

t t

t

t t

t

t

USING THE LIMIT LAWS Example 6

Thus,

Some limits are best calculated by first finding

the left- and right-hand limits.

The following theorem states that a two-sided

limit exists if and only if both the one-sided

limits exist and are equal.

if and only if

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

USING THE LIMIT LAWS

lim ( )x a

f x L

lim ( ) lim ( )x a x a

f x L f x

Theorem 1

Show that

Recall that:

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

Since |x| = -x for x < 0, we have:

Therefore, by Theorem 1, .

0lim 0.x

x

USING THE LIMIT LAWS Example 7

0

0

x if xx

x if x

0 0lim lim 0x x

x x

0 0lim lim( ) 0x x

x x

0lim 0x

x

The result looks plausible from

the figure.

USING THE LIMIT LAWS Example 7

Prove that does not exist.

Since the right- and left-hand limits are different, it follows from Theorem 1 that does not exist.

USING THE LIMIT LAWS Example 8

0limx

x

x

0 0 0

0 0 0

lim lim lim 1 1

lim lim lim( 1) 1

x x x

x x x

x x

x xx x

x x

0limx

x

x

The graph of the function

is shown in the figure.

It supports the one-sided limits that we

found.

USING THE LIMIT LAWS Example 8

( ) | | /f x x x

If

determine whether exists.

Since for x > 4, we have:

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

4 4( )

8 2 4

x if xf x

x if x

4lim ( )x

f x

USING THE LIMIT LAWS Example 9

( ) 4f x x

4 4lim ( ) lim 4

4 4 0

x xf x x

4 4lim ( ) lim (8 2 )

8 2 4 0x x

f x x

The right- and left-hand limits are equal.

Thus, the limit exists and .

4lim ( ) 0x

f x

USING THE LIMIT LAWS Example 9

The graph of f is shown in the

figure.

USING THE LIMIT LAWS Example 9

The greatest integer function is defined

by = the largest integer that is less

than or equal to x.

For instance, , , , , and .

The greatest integer function is sometimes called the floor function.

x

4 4

2

1

4.8 4

3

1

2

1

GREATEST INTEGER FUNCTION

Show that does not exist.

The graph of the greatest integer function is shown in the figure.

USING THE LIMIT LAWS Example 10

limx 3

x

Since for , we have:

Since for , we have:

As these one-sided limits are not equal, does not exist by Theorem 1.

x 2 2 3x

limx 3

x limx 3

2 2

limx 3

x

x 3 3 4x

3 3

lim lim 3 3

x x

x

USING THE LIMIT LAWS Example 10

The next two theorems

give two additional properties

of limits.

USING THE LIMIT LAWS

If when x is near a (except

possibly at a) and the limits of f and g

both exist as x approaches a, then

( ) ( )f x g x

lim ( ) lim ( )x a x a

f x g x

USING THE LIMIT LAWS Theorem 2

The Squeeze Theorem states that, if

when x is near (except

possibly at a ) and ,

then

The Squeeze Theorem is sometimes called the Sandwich Theorem or the Pinching Theorem.

( ) ( ) ( )f x g x h x lim ( ) lim ( )x a x a

f x h x L

lim ( )x a

g x L

USING THE LIMIT LAWS Theorem 3

The theorem is illustrated by the figure.

It states that, if g(x) is squeezed between f(x) and h(x) near a and if f and h have the same limit L at a, then g is forced to have the same limit L at a.

THE SQUEEZE THEOREM

Show that

Note that we cannot use

This is because does not exist.

2

0

1lim sin 0.x

xx

USING THE LIMIT LAWS Example 11

2 2

0 0 0

1 1lim sin lim limsinx x x

x xx x

0limsin(1/ )x

x

However, since , we have:

This is illustrated by the figure.

11 sin 1

x

2 2 21sinx x x

x

USING THE LIMIT LAWS Example 11

We know that: and

Taking f(x) = -x2, , and h(x) = x2 in the Squeeze Theorem, we obtain:

2 2

0 0lim 0 lim( ) 0x x

x x

2( ) sin 1g x x x

2

0

1lim sin 0x

xx

USING THE LIMIT LAWS Example 11