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Calculate limx’3 4x 3 ’ 7x 2 + 5x ’ 9.
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.
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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).
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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

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