SOLUTION

We may apply Theorem 2.1(a) repeatedly to see that

lim 4x 3 ’ 7x 2 + 5x ’ 9 = lim 4x 3 ’ lim 7x 2 + lim 5x ’ lim 9. (—)

x’3 x’3 x’3 x’3 x’3

We next observe that limx’3 x = 3. This assertion is self-evident, for when x

is near to 3 then x is near to 3. Applying Theorem 2.1(d) and Theorem 2.1(b)

repeatedly, we now see that

lim 4x 3 = 4 · lim x · lim x · lim x = 4 · 3 · 3 · 3 = 108.

x’3 x’3 x’3 x’3

Also

lim 7x 2 = 7 · lim x · lim x = 7 · 3 · 3 = 63,

x’3 x’3 x’3

lim 5x = 5 · lim x = 5 · 3 = 15.

x’3 x’3

Of course limx’3 9 = 9.

Putting all this information into equation (—) gives

lim 4x 3 ’ 7x 2 + 5x ’ 9 = 108 ’ 63 + 15 ’ 9 = 51.

x’3

EXAMPLE 2.6

Use the Pinching Theorem to analyze the limit

lim x sin x.

x’0

SOLUTION

We observe that

’|x| ≡ g(x) ¤ f (x) = x sin x ¤ h(x) ≡ |x|.

Thus we may apply the Pinching Theorem. Obviously

lim g(x) = lim h(x) = 0.

x’0 x’0

We conclude that limx’0 f (x) = 0.

64 CHAPTER 2 Foundations of Calculus

EXAMPLE 2.7

Analyze the limit

x2 + 4

lim .

x+2

x’’2

SOLUTION

The denominator tends to 0 while the numerator does not. According to

Theorem 2.3, the limit cannot exist.

You Try It: Use the Pinching Theorem to calculate limx’0 x 2 sin x.

x2

You Try It: What can you say about limx’’1 ?

x2 ’ 1

2.3 Continuity

Let f be a function whose domain contains the interval (a, b). Assume that c is a

point of (a, b). We say that the function f is continuous at c if

lim f (x) = f (c).

x’c

Conceptually, f is continuous at c if the expected value of f at c equals the actual

value of f at c.

EXAMPLE 2.8

Is the function

2x 2 ’ x if x < 2

f (x) =

if x ≥ 2

3x

continuous at x = 2?

SOLUTION

We easily check that limx’2 f (x) = 6. Also the actual value of f at 2, given

by the second part of the formula, is equal to 6. By the de¬nition of continuity,

we may conclude that f is continuous at x = 2. See Fig. 2.5.

EXAMPLE 2.9

Where is the function

±

1 if x < 4

g(x) = x ’ 3

2x + 3 if x ≥ 4

continuous?

65

CHAPTER 2 Foundations of Calculus

Fig. 2.5

SOLUTION

If x < 3 then the function is plainly continuous. The function is unde¬ned

at x = 3 so we may not even speak of continuity at x = 3. The function is also

obviously continuous for 3 < x < 4. At x = 4 the limit of g does not exist”it

is 1 from the left and 11 from the right. So the function is not continuous (we

sometimes say that it is discontinuous) at x = 4. By inspection, the function is

continuous for x > 4.

You Try It: Discuss continuity of the function

±

x ’ x 2 if x < ’2

g(x) = 10 if x = ’2

’5x if x > ’2

We note that Theorem 2.1 guarantees that the collection of continuous functions

is closed under addition, subtraction, multiplication, division (as long as we do not

divide by 0), and scalar multiplication.

Math Note: If f —¦ g makes sense, if limx’c g(x) = , and if lims’ f (s) = m,

then it does not necessarily follow that limx’c f —¦ g(x) = m. [We invite the reader

to ¬nd an example.] One must assume, in addition, that f is continuous at . This

point will come up from time to time in our later studies.

In the next section we will learn the concept of the derivative. It will turn out

that a function that possesses the derivative is also continuous.

66 CHAPTER 2 Foundations of Calculus

2.4 The Derivative

Suppose that f is a function whose domain contains the interval (a, b). Let c be a

point of (a, b). If the limit

f (c + h) ’ f (c)

lim (—)

h

h’0

exists then we say that f is differentiable at c and we call the limit the derivative

of f at c.

EXAMPLE 2.10

Is the function f (x) = x 2 + x differentiable at x = 2? If it is, calculate the

derivative.

SOLUTION

We calculate the limit (—), with the role of c played by 2:

f (2 + h) ’ f (2) [(2 + h)2 + (2 + h)] ’ [22 + 2]

= lim

lim

h h

h’0 h’0

[(4 + 4h + h2 ) + (2 + h)] ’ [6]

= lim

h

h’0

5h + h2

= lim

h

h’0

= lim 5 + h

h’0

= 5.

We see that the required limit (—) exists, and that it equals 5. Thus the function

f (x) = x 2 + x is differentiable at x = 2, and the value of the derivative is 5.

Math Note: When the derivative of a function f exists at a point c, then we denote

the derivative either by f (c) or by (d/dx)f (c) = (df /dx)(c). In some contexts

(e.g., physics) the notation f™(c) is used. In the last example, we calculated that

f (2) = 5.

The importance of the derivative is two-fold: it can be interpreted as rate of

change and it can be interpreted as the slope. Let us now consider both of these

ideas.

Suppose that •(t) represents the position (in inches or feet or some other standard

unit) of a moving body at time t. At time 0 the body is at •(0), at time 3 the body is

at •(3), and so forth. Imagine that we want to determine the instantaneous velocity

of the body at time t = c. What could this mean? One reasonable interpretation

is that we can calculate the average velocity over a small interval at c, and let the

67

CHAPTER 2 Foundations of Calculus

length of that interval shrink to zero to determine the instantaneous velocity. To

carry out this program, imagine a short interval [c, c + h]. The average velocity of

the moving body over that interval is

•(c + h) ’ •(c)

vav ≡ .

h

This is a familiar expression (see (—)).As we let h ’ 0, we know that this expression

tends to the derivative of • at c. On the other hand, it is reasonable to declare this

limit to be the instantaneous velocity. We have discovered the following important

rule:

Let • be a differentiable function on an interval (a, b). Suppose that •(t )

represents the position of a moving body. Let c ∈ (a, b). Then

• (c ) = instantaneous velocity of the moving body at c.

Now let us consider slope. Look at the graph of the function y = f (x) in Fig. 2.6.

We wish to determine the “slope” of the graph at the point x = c. This is the same

as determining the slope of the tangent line to the graph of f at x = c, where the

tangent line is the line that best approximates the graph at that point. See Fig. 2.7.

What could this mean? After all, it takes two points to determine the slope of a line,

yet we are only given the point (c, f (c)) on the graph. One reasonable interpretation

of the slope at (c, f (c)) is that it is the limit of the slopes of secant lines determined

Fig. 2.6

Fig. 2.7

68 CHAPTER 2 Foundations of Calculus

by (c, f (c)) and nearby points (c + h, f (c + h)). See Fig. 2.8. Let us calculate this

limit:

f (c + h) ’ f (c) f (c + h) ’ f (c)

= lim

lim .

(c + h) ’ c h

h’0 h’0

Fig. 2.8

We know that this last limit (the same as (—)) is the derivative of f at c. We have

Y

learned the following:

FL

Let f be a differentiable function on an interval (a, b). Let c ∈ (a, b). Then

the slope of the tangent line to the graph of f at c is f (c ).

AM

EXAMPLE 2.11

Calculate the instantaneous velocity at time t = 5 of an automobile whose

position at time t seconds is given by g(t) = t 3 + 4t 2 + 10 feet.

TE

SOLUTION

We know that the required instantaneous velocity is g (5). We calculate

g(5 + h) ’ g(5)

g (5) = lim

h

h’0

[(5 + h)3 + 4(5 + h)2 + 10] ’ [53 + 4 · 52 + 10]

= lim

h

h’0

((125 + 75h + 15h2 + h3 ) + 4 · (25 + 10h + h2 ) + 10)

= lim

h

h’0

(125 + 100 + 10)

’

h

115h + 19h2 + h3

= lim

h

h’0

= lim 115 + 19h + h2

h’0

= 115.

69

CHAPTER 2 Foundations of Calculus

We conclude that the instantaneous velocity of the moving body at time t = 5

is g (5) = 115 ft/sec.

Math Note: Since position (or distance) is measured in feet, and time in seconds,

then we measure velocity in feet per second.

EXAMPLE 2.12

Calculate the slope of the tangent line to the graph of y = f (x) = x 3 ’ 3x

at x = ’2. Write the equation of the tangent line. Draw a ¬gure illustrating

these ideas.

SOLUTION

We know that the desired slope is equal to f (’2). We calculate

f (’2 + h) ’ f (’2)

f (’2) = lim

h

h’0

[(’2 + h)3 ’ 3(’2 + h)] ’ [(’2)3 ’ 3(’2)]

= lim

h

h’0

[(’8 + 12h ’ 6h2 + h3 ) + (6 ’ 3h)] + [2]

= lim

h

h’0

3 ’ 6h2 + 9h

h

= lim

h

h’0

= lim h2 ’ 6h + 9

h’0

= 9.

We conclude that the slope of the tangent line to the graph of y = x 3 ’ 3x at

x = ’2 is 9. The tangent line passes through (’2, f (’2)) = (’2, ’2) and

has slope 9. Thus it has equation

y ’ (’2) = 9(x ’ (’2)).

The graph of the function and the tangent line are exhibited in Fig. 2.9.

You Try It: Calculate the tangent line to the graph of f (x) = 4x 2 ’ 5x + 2 at the

point where x = 2.

EXAMPLE 2.13

A rubber balloon is losing air steadily. At time t minutes the balloon contains

75 ’ 10t 2 + t cubic inches of air. What is the rate of loss of air in the balloon

at time t = 1?

70 CHAPTER 2 Foundations of Calculus

x

Fig. 2.9

SOLUTION

Let ψ(t) = 75 ’ 10t 2 + t. Of course the rate of loss of air is given by ψ (1).

We therefore calculate

ψ(1 + h) ’ ψ(1)

ψ (1) = lim

h

h’0

[75 ’ 10(1 + h)2 + (1 + h)] ’ [75 ’ 10 · 12 + 1]

= lim

h

h’0

[75 ’ (10 + 20h + 10h2 ) + (1 + h)] ’ [66]

= lim

h

h’0

’19h ’ 10h2

= lim

h

h’0

= lim ’19 ’ 10h

h’0

= ’19.

In conclusion, the rate of air loss in the balloon at time t = 1 is ψ (1) =

’19 ft 3 /sec. Observe that the negative sign in this answer indicates that the

change is negative, i.e., that the quantity is decreasing.

You Try It: The amount of water in a leaky tank is given by W (t) = 50 ’ 5t 2 + t

gallons. What is the rate of leakage of the water at time t = 2?

Math Note: We have noted that the derivative may be used to describe a rate of

change and also to denote the slope of the tangent line to a graph. These are really

two different manifestations of the same thing, for a slope is the rate of change of

rise with respect to run (see Section 1.4 on the slope of a line).

71

CHAPTER 2 Foundations of Calculus

2.5 Rules for Calculating Derivatives

Calculus is a powerful tool, for much of the physical world that we wish to analyze

is best understood in terms of rates of change. It becomes even more powerful when

we can ¬nd some simple rules that enable us to calculate derivatives quickly and

easily. This section is devoted to that topic.

I Derivative of a Sum [The Sum Rule]: We calculate the derivative of a sum

(or difference) by

(f (x) ± g(x)) = f (x) ± g (x).

In our many examples, we have used this fact implicitly. We are now just

enunciating it formally.

II Derivative of a Product [The Product Rule]: We calculate the derivative

of a product by

[f (x) · g(x)] = f (x) · g(x) + f (x) · g (x).

We urge the reader to test this formula on functions that we have worked with

before. It has a surprising form. Note in particular that it is not the case that

[f (x) · g(x)] = f (x) · g (x).

III Derivative of a Quotient [The Quotient Rule]: We calculate the derivative

of a quotient by

g(x) · f (x) ’ f (x) · g (x)

f (x)

= .

g 2 (x)

g(x)

In fact one can derive this new formula by applying the product formula to

g(x) · [f (x)/g(x)]. We leave the details for the interested reader.

IV Derivative of a Composition [The Chain Rule]: We calculate the

derivative of a composition by

[f —¦ g(x)] = f (g(x)) · g (x).

To make optimum use of these four new formulas, we need a library of functions

to which to apply them.

A Derivatives of Powers of x: If f (x) = x k then f (x) = k · x k’1 , where

k ∈ {0, 1, 2, . . .}.

Math Note: If you glance back at the examples we have done, you will notice that

we have already calculated that the derivative of x is 1, the derivative of x 2 is 2x,

72 CHAPTER 2 Foundations of Calculus

and the derivative of x 3 is 3x 2 . The rule just enunciated is a generalization of these

facts, and is established in just the same way.

B Derivatives of Trigonometric Functions: The rules for differentiating sine

and cosine are simple and elegant:

d

sin x = cos x.

1.

dx

d

cos x = ’ sin x.

2.

dx