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Unformatted text preview: 5E02(pp 064073) 1/17/06 1:24 PM Page 64 CHAPTER 2 The idea of a limit is
illustrated by secant lines
approaching a tangent line. L imits and Rates of Change 5E02(pp 064073) 1/17/06 1:25 PM Page 65 In A Preview of Calculus (page 2) we saw how the idea of
a limit underlies the various branches of calculus. It is therefore appropriate to begin our study of calculus by investigating limits and their properties.  2.1 The Tangent and Velocity Problems
In this section we see how limits arise when we attempt to ﬁnd the tangent to a curve or
the velocity of an object. The Tangent Problem
The word tangent is derived from the Latin word tangens, which means “touching.” Thus,
a tangent to a curve is a line that touches the curve. In other words, a tangent line should
have the same direction as the curve at the point of contact. How can this idea be made
precise?
For a circle we could simply follow Euclid and say that a tangent is a line that intersects
the circle once and only once as in Figure 1(a). For more complicated curves this deﬁnition is inadequate. Figure l(b) shows two lines l and t passing through a point P on a curve
C. The line l intersects C only once, but it certainly does not look like what we think of as
a tangent. The line t, on the other hand, looks like a tangent but it intersects C twice. Locate tangents interactively and explore
them numerically.
Resources / Module 1
/ Tangents
/ What Is a Tangent? t P
t C l
FIGURE 1 (a) y Q { x, ≈} FIGURE 2 t P (1, 1)
0 To be speciﬁc, let’s look at the problem of trying to ﬁnd a tangent line t to the parabola
x 2 in the following example. EXAMPLE 1 Find an equation of the tangent line to the parabola y y y=≈ (b) x x 2 at the point P 1, 1 . SOLUTION We will be able to ﬁnd an equation of the tangent line t as soon as we know its
slope m. The difﬁculty is that we know only one point, P, on t, whereas we need two
points to compute the slope. But observe that we can compute an approximation to m by
choosing a nearby point Q x, x 2 on the parabola (as in Figure 2) and computing the
slope mPQ of the secant line PQ.
We choose x 1 so that Q P. Then mPQ x2
x 1
1 65 5E02(pp 064073) 66 ❙❙❙❙ 1/17/06 1:25 PM Page 66 CHAPTER 2 LIMITS AND RATES OF CHANGE For instance, for the point Q 1.5, 2.25 we have
mPQ
x 3
2.5
2.1
2.01
2.001 lim mPQ Q lP x 1.25
0.5 2.5 and m lim xl1 x2
x 1
1 2 Assuming that the slope of the tangent line is indeed 2, we use the pointslope form
of the equation of a line (see Appendix B) to write the equation of the tangent line
through 1, 1 as mPQ 0
0.5
0.9
0.99
0.999 1
1 The tables in the margin show the values of mPQ for several values of x close to 1. The
closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mPQ is to 2.
This suggests that the slope of the tangent line t should be m 2.
We say that the slope of the tangent line is the limit of the slopes of the secant lines,
and we express this symbolically by writing mPQ 2
1.5
1.1
1.01
1.001 2.25
1.5 1
1.5
1.9
1.99
1.999 y 1 2x 1 or y 2x 1 Figure 3 illustrates the limiting process that occurs in this example. As Q approaches
P along the parabola, the corresponding secant lines rotate about P and approach the
tangent line t. y y y Q
t t t Q
Q
P P 0 P 0 x x 0 x Q approaches P from the right
y y y t t Q P t P P Q
0 x Q
0 x 0 x Q approaches P from the left
FIGURE 3
In Module 2.1 you can see how the
process in Figure 3 works for ﬁve
additional functions. Many functions that occur in science are not described by explicit equations; they are
deﬁned by experimental data. The next example shows how to estimate the slope of the
tangent line to the graph of such a function. 5E02(pp 064073) 1/17/06 1:25 PM Page 67 SECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS t 67 E XAMPLE 2 The ﬂash unit on a camera operates by storing charge on a capacitor and
releasing it suddenly when the ﬂash is set off. The data at the left describe the charge Q
remaining on the capacitor (measured in microcoulombs) at time t (measured in seconds
after the ﬂash goes off ). Use the data to draw the graph of this function and estimate the
slope of the tangent line at the point where t 0.04. [ Note: The slope of the tangent line
represents the electric current ﬂowing from the capacitor to the ﬂash bulb (measured in
microamperes).] Q 0.00
0.02
0.04
0.06
0.08
0.10 ❙❙❙❙ 100.00
81.87
67.03
54.88
44.93
36.76 SOLUTION In Figure 4 we plot the given data and use them to sketch a curve that approximates the graph of the function.
Q
100
90
80 A
P 70
60
50
0 B C 0.02 0.04 0.06 0.08 0.1 t F IGURE 4 Given the points P 0.04, 67.03 and R 0.00, 100.00 on the graph, we ﬁnd that the
slope of the secant line PR is
100.00
0.00 mPR
R
(0.00, 100.00)
(0.02, 81.87)
(0.06, 54.88)
(0.08, 44.93)
(0.10, 36.76) mPR
824.25
742.00
607.50
552.50
504.50 67.03
0.04 824.25 The table at the left shows the results of similar calculations for the slopes of other
secant lines. From this table we would expect the slope of the tangent line at t 0.04 to
lie somewhere between 742 and 607.5. In fact, the average of the slopes of the two
closest secant lines is
1
2 742 607.5 674.75 So, by this method, we estimate the slope of the tangent line to be 675.
Another method is to draw an approximation to the tangent line at P and measure the
sides of the triangle ABC, as in Figure 4. This gives an estimate of the slope of the tangent line as  The physical meaning of the answer in
Example 2 is that the electric current ﬂowing
from the capacitor to the ﬂash bulb after
0.04 second is about –670 microamperes. AB
BC 80.4
0.06 53.6
0.02 670 The Velocity Problem
If you watch the speedometer of a car as you travel in city trafﬁc, you see that the needle
doesn’t stay still for very long; that is, the velocity of the car is not constant. We assume
from watching the speedometer that the car has a deﬁnite velocity at each moment, but how
is the “instantaneous” velocity deﬁned? Let’s investigate the example of a falling ball. 5E02(pp 064073) 68 ❙❙❙❙ 1/17/06 1:25 PM Page 68 CHAPTER 2 LIMITS AND RATES OF CHANGE EXAMPLE 3 Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 5 seconds.
SOLUTION Through experiments carried out four centuries ago, Galileo discovered that the
distance fallen by any freely falling body is proportional to the square of the time it has
been falling. (This model for free fall neglects air resistance.) If the distance fallen after t
seconds is denoted by s t and measured in meters, then Galileo’s law is expressed by the
equation 4.9t 2 st The difﬁculty in ﬁnding the velocity after 5 s is that we are dealing with a single instant
of time t 5 , so no time interval is involved. However, we can approximate the desired
quantity by computing the average velocity over the brief time interval of a tenth of a
second from t 5 to t 5.1:
average velocity distance traveled
time elapsed
s 5.1 The CN Tower in Toronto is currently the tallest
freestanding building in the world. s5
0.1 4.9 5.1 2 4.9 5 2 0.1 49.49 m s The following table shows the results of similar calculations of the average velocity over
successively smaller time periods.
Time interval
5
5
5
5
5 t
t
t
t
t Average velocity (m s) 6
5.1
5.05
5.01
5.001 53.9
49.49
49.245
49.049
49.0049 It appears that as we shorten the time period, the average velocity is becoming closer to
49 m s. The instantaneous velocity when t 5 is deﬁned to be the limiting value of
these average velocities over shorter and shorter time periods that start at t 5. Thus,
the (instantaneous) velocity after 5 s is
v 49 m s You may have the feeling that the calculations used in solving this problem are very
similar to those used earlier in this section to ﬁnd tangents. In fact, there is a close connection between the tangent problem and the problem of ﬁnding velocities. If we draw the
graph of the distance function of the ball (as in Figure 5) and we consider the points
P a, 4.9a 2 and Q a h, 4.9 a h 2 on the graph, then the slope of the secant line
PQ is
mPQ 4.9 a
a h2
h 4.9a 2
a 5E02(pp 064073) 1/17/06 1:26 PM Page 69 S ECTION 2.1 THE TANGENT AND VELOCITY PROBLEMS ❙❙❙❙ 69 which is the same as the average velocity over the time interval a, a h . Therefore, the
velocity at time t a (the limit of these average velocities as h approaches 0) must be
equal to the slope of the tangent line at P (the limit of the slopes of the secant lines).
s s s=4.9t@ s=4.9t@ Q
slope of secant line
average velocity a 0 slope of tangent
instantaneous velocity P P a+h t 0 a t FIGURE 5 Examples 1 and 3 show that in order to solve tangent and velocity problems we must
be able to ﬁnd limits. After studying methods for computing limits in the next four sections,
we will return to the problems of ﬁnding tangents and velocities in Section 2.6.  2.1 Exercises 1. A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show
the volume V of water remaining in the tank (in gallons) after
t minutes.
t (min) 5 10 15 20 25 30 V (gal) 694 444 250 111 28 0 (a) If P is the point 15, 250 on the graph of V, ﬁnd the slopes
of the secant lines PQ when Q is the point on the graph
with t 5, 10, 20, 25, and 30.
(b) Estimate the slope of the tangent line at P by averaging the
slopes of two secant lines.
(c) Use a graph of the function to estimate the slope of the
tangent line at P. (This slope represents the rate at which the
water is ﬂowing from the tank after 15 minutes.)
2. A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t minutes. When the data in the table are graphed, the slope of the
tangent line represents the heart rate in beats per minute.
t (min)
Heartbeats 36 38 40 42 44 2530 2661 2806 2948 3080 The monitor estimates this value by calculating the slope
of a secant line. Use the data to estimate the patient’s heart rate after 42 minutes using the secant line between the points with
the given values of t.
(a) t 36 and t 42
(b) t 38 and t 42
(c) t 40 and t 42
(d) t 42 and t 44
What are your conclusions?
3. The point P (1, 2 ) lies on the curve y 1
x 1 x.
(a) If Q is the point x, x 1 x , use your calculator to ﬁnd
the slope of the secant line PQ (correct to six decimal
places) for the following values of x :
(i) 0.5
(ii) 0.9
(iii) 0.99
(iv) 0.999
(v) 1.5
(vi) 1.1
(vii) 1.01
(viii) 1.001
(b) Using the results of part (a), guess the value of the slope of
the tangent line to the curve at P (1, 1 ).
2
(c) Using the slope from part (b), ﬁnd an equation of the
tangent line to the curve at P (1, 1 ).
2 4. The point P 5, 3 lies on the curve y sx 4.
(a) If Q is the point ( x, sx 4 ), use your calculator to ﬁnd
the slope of the secant line PQ (correct to six decimal
places) for the following values of x :
(i) 4.5
(ii) 4.9
(iii) 4.99
(iv) 4.999
(v) 5.5
(vi) 5.1
(vii) 5.01
(viii) 5.001
(b) Using the results of part (a), guess the value of the slope of
the tangent line to the curve at P 5, 3 . 5E02(pp 064073) 70 ❙❙❙❙ 1/17/06 1:26 PM Page 70 CHAPTER 2 LIMITS AND RATES OF CHANGE (c) Draw the graph of s as a function of t and draw the secant
lines whose slopes are the average velocities found in
part (a).
(d) Draw the tangent line whose slope is the instantaneous
velocity from part (b). (c) Using the slope from part (b), ﬁnd an equation of the
tangent line to the curve at P 5, 3 .
(d) Sketch the curve, two of the secant lines, and the tangent
line.
5. If a ball is thrown into the air with a velocity of 40 ft s, its height in feet after t seconds is given by y 40 t 16 t 2.
(a) Find the average velocity for the time period beginning
when t 2 and lasting
(i) 0.5 second
(ii) 0.1 second
(iii) 0.05 second
(iv) 0.01 second
(b) Find the instantaneous velocity when t 2. 8. The position of a car is given by the values in the table.
t (seconds) 1 2 3 4 5 s (feet) 0 10 32 70 119 178 (a) Find the average velocity for the time period beginning
when t 2 and lasting
(i) 3 seconds
(ii) 2 seconds
(iii) 1 second
(b) Use the graph of s as a function of t to estimate the instantaneous velocity when t 2. 6. If an arrow is shot upward on the moon with a velocity of 58 m s, its height in meters after t seconds is given by
h 58t 0.83 t 2.
(a) Find the average velocity over the given time intervals:
(i) [1, 2]
(ii) [1, 1.5]
(iii) [1, 1.1]
(iv) [1, 1.01]
(v) [1, 1.001]
(b) Find the instantaneous velocity after one second. 9. The point P 1, 0 lies on the curve y 7. The displacement (in feet) of a certain particle moving in a straight line is given by s t 3 6, where t is measured in
seconds.
(a) Find the average velocity over the following time periods:
(i) [1, 3]
(ii) [1, 2]
(iii) [1, 1.5]
(iv) [1, 1.1]
(b) Find the instantaneous velocity when t 1.  2.2 0 ; sin 10 x .
(a) If Q is the point x, sin 10 x , ﬁnd the slope of the secant
line PQ (correct to four decimal places) for x 2, 1.5, 1.4,
1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. Do the slopes
appear to be approaching a limit?
(b) Use a graph of the curve to explain why the slopes of the
secant lines in part (a) are not close to the slope of the
tangent line at P.
(c) By choosing appropriate secant lines, estimate the slope of
the tangent line at P. The Limit of a Function
Having seen in the preceding section how limits arise when we want to ﬁnd the tangent to
a curve or the velocity of an object, we now turn our attention to limits in general and
numerical and graphical methods for computing them.
Let’s investigate the behavior of the function f deﬁned by f x
x 2 x 2 for values of x near 2. The following table gives values of f x for values of x close to 2, but not
equal to 2. y x
ƒ
approaches
4. y=≈ x+2 4 0 2 As x approaches 2,
FIGURE 1 fx x fx 1.0
1.5
1.8
1.9
1.95
1.99
1.995
1.999 2.000000
2.750000
3.440000
3.710000
3.852500
3.970100
3.985025
3.997001 3.0
2.5
2.2
2.1
2.05
2.01
2.005
2.001 8.000000
5.750000
4.640000
4.310000
4.152500
4.030100
4.015025
4.003001 x From the table and the graph of f (a parabola) shown in Figure 1 we see that when x is
close to 2 (on either side of 2), f x is close to 4. In fact, it appears that we can make the 5E02(pp 064073) 1/17/06 1:26 PM Page 71 S ECTION 2.2 THE LIMIT OF A FUNCTION ❙❙❙❙ 71 values of f x as close as we like to 4 by taking x sufﬁciently close to 2. We express this
by saying “the limit of the function f x
x 2 x 2 as x approaches 2 is equal to 4.”
The notation for this is
lim x 2
x l2 x 2 4 In general, we use the following notation.
1 Definition We write lim f x xla and say L “the limit of f x , as x approaches a, equals L” if we can make the values of f x arbitrarily close to L (as close to L as we like)
by taking x to be sufﬁciently close to a (on either side of a) but not equal to a.
Roughly speaking, this says that the values of f x get closer and closer to the number
L as x gets closer and closer to the number a (from either side of a) but x a. A more precise deﬁnition will be given in Section 2.4.
An alternative notation for
lim f x xla is f x lL as L
xla which is usually read “ f x approaches L as x approaches a.”
Notice the phrase “but x a” in the deﬁnition of limit. This means that in ﬁnding the
limit of f x as x approaches a, we never consider x a. In fact, f x need not even be
deﬁned when x a. The only thing that matters is how f is deﬁned near a.
Figure 2 shows the graphs of three functions. Note that in part (c), f a is not deﬁned
and in part (b), f a
L. But in each case, regardless of what happens at a, it is true that
lim x l a f x
L.
y y y L L L 0 a x (a) 0 a 0 x (b) a x (c) FIGURE 2 lim ƒ=L in all three cases
xa EXAMPLE 1 Guess the value of lim
x l1 x
x2 1
.
1 x 1 x 2 1 is not deﬁned when x 1,
but that doesn’t matter because the deﬁnition of lim x l a f x says that we consider values
SOLUTION Notice that the function f x 5E02(pp 064073) 72 ❙❙❙❙
x 1/17/06 1:27 PM Page 72 CHAPTER 2 LIMITS AND RATES OF CHANGE 1 0.5
0.9
0.99
0.999
0.9999 of x that are close to a but not equal to a. The tables at the left give values of f x
(correct to six decimal places) for values of x that approach 1 (but are not equal to 1).
On the basis of the values in the tables, we make the guess that fx
0.666667
0.526316
0.502513
0.500250
0.500025 lim
x l1 x
x2 1
1 0.5 Example 1 is illustrated by the graph of f in Figure 3. Now let’s change f slightly by
giving it the value 2 when x 1 and calling the resulting function t :
x 1 1.5
1.1
1.01
1.001
1.0001 fx
0.400000
0.476190
0.497512
0.499750
0.499975 x
x2 t (x) 1
1 1 if x 2 if x 1 This new function t still has the same limit as x approaches 1 (see Figure 4).
y y
2 y= x1
≈1 y=© 0.5
0 0.5 1 x 0 FIGURE 3 1 x FIGURE 4 EXAMPLE 2 Estimate the value of lim st 2 tl0 9
t 3 2 . SOLUTION The table lists values of the function for several values of t near 0. t
1.0
0.5
0.1
0.05
0.01 t
0.0005
0.0001
0.00005
0.00001 st 2 9
t2 0.16800
0.20000
0.00000
0.00000 3 st 2 9
t 3 2 0.16228
0.16553
0.16662
0.16666
0.16667 As t approaches 0, the values of the function seem to approach 0.1666666 . . . and so we
guess that
1
st 2 9 3
lim
2
tl0
t
6
In Example 2 what would have happened if we had taken even smaller values of t ? The
table in the margin shows the results from one calculator; you can see that something
strange seems to be happening. 5E02(pp 064073) 1/17/06 1:27 PM Page 73 SECTION 2.2 THE LIMIT OF A FUNCTION ❙❙❙❙ 73 If you try these calculations on your own calculator you might get different values, but
eventually you will get the value 0 if you make t sufﬁciently small. Does this mean that
the answer is really 0 instead of 1 ? No, the value of the limit is 1 , as we will show in the
6
6
 next section. The problem is that the calculator gave false values because st 2 9 is very
close to 3 when t is small. (In fact, when t is sufﬁciently small, a calculator’s value for
 For a further explanation of why calculators
st 2 9 is 3.000 . . . to as many digits as the calculator is capable of carrying.)
sometimes give false values, see the web site
Something similar happens when we try to graph the function
www.stewartcalculus.com
Click on A dditional Topics and then on L ies
My Calculator and Computer Told Me . In
particular, see the section called The Perils of
Subtraction. st 2 ft 9
t 3 2 of Example 2 on a graphing calculator or computer. Parts (a) and (b) of Figure 5 show quite
accurate graphs of f , and when we use the trace mode (if available) we can estimate easily that the limit is about 1. But if we zoom in too far, as in parts (c) and (d), then we get
6
inaccurate graphs, again because of problems with subtraction. 0.2 0.2 0.1 0.1 (a) _5, 5 by _ 0.1, 0.3 (b) _ 0.1, 0.1 by _ 0.1, 0.3 (c) _10– ^, 10– ^ by _ 0.1, 0.3 (d) _10– &, 10– & by _ 0.1, 0.3 F IGURE 5 EXAMPLE 3 Guess the value of lim xl0 sin x
.
x SOLUTION The function f x
sin x x is not deﬁned when x 0. Using a calculator
(and remembering that, if x
, sin x means the sine of the angle whose radian measure is x), we construct the following table of values correct to eight decimal places.
From the table and the graph in Figure 6 we guess that lim xl0 sin x
x 1 This guess is in fact correct, as will be proved in Chapter 3 using a geometric argument. x
1.0
0.5
0.4
0.3
0.2
0.1
0.05
0.01
0.005
0.001 sin x
x
0.84147098
0.95885108
0.97354586
0.98506736
0.99334665
0.99833417
0.99958339
0.99998333
0.99999583
0.99999983 y
1 _1 FIGURE 6 y= 0 1 sin x
x x 5E02(pp 074085) 74 ❙❙❙❙ 1/17/06 1:16 PM Page 74 CHAPTER 2 LIMITS AND RATES OF CHANGE  COMPUTER ALGEBRA SYSTEMS
Computer algebra systems (CAS) have commands
that compute limits. In order to avoid the types
of pitfalls demonstrated in Examples 2, 4, and 5,
they don’t ﬁnd limits by numerical experimentation. Instead, they use more sophisticated
techniques such as computing inﬁnite series. If
you have access to a CAS, use the limit command
to compute the limits in the examples of this
section and to check your answers in the exercises of this chapter. EXAMPLE 4 Investigate lim sin
xl0 x . SOLUTION Again the function f x
sin
for some small values of x, we get f1 sin f ( 1)
3 sin 3 f 0.1 x is undeﬁned at 0. Evaluating the function
f (1)
2 0 sin 10 Similarly, f 0.001
f 0.0001
tempted to guess that 0 sin 2 0 f (1)
4 0 sin 4 0 f 0.01 sin 100 0 0. On the basis of this information we might be lim sin xl0 0 x  but this time our guess is wrong. Note that although f 1 n sin n
0 for any integer
1 for inﬁnitely many values of x that approach 0. [In fact, n, it is also true that f x
sin x
1 when x
and, solving for x, we get x 2 2 4n 2n 1 .] The graph of f is given in Figure 7.
y y=sin(π/x) 1
Listen to the sound of this function trying to
approach a limit.
Resources / Module 2
/ Basics of Limits
/ Sound of a Limit that Does Not Exist _1
1 x _1 F IGURE 7 Module 2.2 helps you explore limits at
points where graphs exhibit unusual
behavior. The dashed lines indicate that the values of sin x oscillate between 1 and 1
inﬁnitely often as x approaches 0 (see Exercise 37). Since the values of f x do not
approach a ﬁxed number as x approaches 0,
lim sin xl0 x x3 cos 5x
10,000 EXAMPLE 5 Find lim x 3
xl0 1
0.5
0.1
0.05
0.01 1.000028
0.124920
0.001088
0.000222
0.000101 x does not exist cos 5x
.
10,000 SOLUTION As before, we construct a table of values. From the table in the margin it
appears that lim x 3 xl0 cos 5x
10,000 0 5E02(pp 074085) 1/17/06 1:16 PM Page 75 S ECTION 2.2 THE LIMIT OF A FUNCTION x
0.005
0.001 x3 lim x 3 xl0 cos 5x
10,000 Later we will see that lim x l 0 cos 5x  0.000100 1
10,000 1; then it follows that the limit is 0.0001. Examples 4 and 5 illustrate some of the pitfalls in guessing the value of a limit. It is
easy to guess the wrong value if we use inappropriate values of x, but it is difﬁcult to know
when to stop calculating values. And, as the discussion after Example 2 shows, sometimes
calculators and computers give the wrong values. In the next two sections, however, we
will develop foolproof methods for calculating limits.
EXAMPLE 6 The Heaviside function H is deﬁned by y
1 FIGURE 8 75 But if we persevere with smaller values of x, the second table suggests that cos 5x
10,000 0.00010009
0.00010000 0 ❙❙❙❙ 0
1 Ht
t if t
if t 0
0 [This function is named after the electrical engineer Oliver Heaviside (1850–1925) and
can be used to describe an electric current that is switched on at time t 0.] Its graph is
shown in Figure 8.
As t approaches 0 from the left, H t approaches 0. As t approaches 0 from the right,
H t approaches 1. There is no single number that H t approaches as t approaches 0.
Therefore, lim t l 0 H t does not exist. OneSided Limits
We noticed in Example 6 that H t approaches 0 as t approaches 0 from the left and H t
approaches 1 as t approaches 0 from the right. We indicate this situation symbolically by
writing
lim H t tl0 0 and lim H t tl0 1 The symbol “t l 0 ” indicates that we consider only values of t that are less than 0.
Likewise, “t l 0 ” indicates that we consider only values of t that are greater than 0.
2 Definition We write lim f x xla L and say the lefthand limit of f x as x approaches a [or the limit of f x as x
approaches a from the left] is equal to L if we can make the values of f x arbitrarily close to L by taking x to be sufﬁciently close to a and x less than a.
Notice that Deﬁnition 2 differs from Deﬁnition 1 only in that we require x to be less
than a. Similarly, if we require that x be greater than a, we get “the righthand limit of
f x as x approaches a is equal to L” and we write
lim f x xla L 5E02(pp 074085) 76 ❙❙❙❙ 1/17/06 1:17 PM Page 76 CHAPTER 2 LIMITS AND RATES OF CHANGE Thus, the symbol “x l a ” means that we consider only x
trated in Figure 9.
y a. These deﬁnitions are illus y L ƒ
0 FIGURE 9 x ƒ L
0 x a a x x (b) lim ƒ=L (a) lim ƒ=L x a+ x a_ By comparing Deﬁnition l with the deﬁnitions of onesided limits, we see that the following is true.
lim f x 3 y L xla if and only if lim f x xla and lim f x xla L EXAMPLE 7 The graph of a function t is shown in Figure 10. Use it to state the values
(if they exist) of the following: 4
3 (a) lim t x (b) lim t x (c) lim t x (d) lim t x y=© (e) lim t x (f) lim t x xl2 xl2 xl5 1
0 L 1 2 3 4 5 x xl2 xl5 xl5 SOLUTION From the graph we see that the values of t x approach 3 as x approaches 2
from the left, but they approach 1 as x approaches 2 from the right. Therefore (a) lim t x F IGURE 10 xl2 and 3 (b) lim t x
xl2 1 (c) Since the left and right limits are different, we conclude from (3) that lim x l 2 t x
does not exist.
The graph also shows that
(d) lim t x
xl5 and 2 (e) lim t x
xl5 2 (f) This time the left and right limits are the same and so, by (3), we have
lim t x xl5 Despite this fact, notice that t 5 2 2. I nfinite Limits
EXAMPLE 8 Find lim xl0 1
if it exists.
x2 SOLUTION As x becomes close to 0, x 2 also becomes close to 0, and 1 x 2 becomes very large. (See the table on the next page.) In fact, it appears from the graph of the function
fx
1 x 2 shown in Figure 11 that the values of f x can be made arbitrarily large 5E02(pp 074085) 1/17/06 1:17 PM Page 77 SECTION 2.2 THE LIMIT OF A FUNCTION 1
0.5
0.2
0.1
0.05
0.01
0.001 77 by taking x close enough to 0. Thus, the values of f x do not approach a number, so
lim x l 0 1 x 2 does not exist. 1
x2 x ❙❙❙❙ y 1
4
25
100
400
10,000
1,000,000 y= 1
≈ x 0 F IGURE 11 To indicate the kind of behavior exhibited in Example 8, we use the notation
lim xl0 1
x2  This does not mean that we are regarding as a number. Nor does it mean that the limit
exists. It simply expresses the particular way in which the limit does not exist: 1 x 2 can be
made as large as we like by taking x close enough to 0.
In general, we write symbolically
lim f x xla to indicate that the values of f x become larger and larger (or “increase without bound”)
as x becomes closer and closer to a. Explore inﬁnite limits interactively.
Resources / Module 2
/ Limits that Are Inﬁnite
/ Examples A and B 4 Definition Let f be a function deﬁned on both sides of a, except possibly at a itself. Then
lim f x xla means that the values of f x can be made arbitrarily large (as large as we please)
by taking x sufﬁciently close to a, but not equal to a. Another notation for lim x l a f x is y fxl y=ƒ Again the symbol
0 a FIGURE 12 lim ƒ=`
xa xla is not a number, but the expression lim x l a f x
“the limit of f x , as x approaches a, is inﬁnity” x x=a as or “ f x becomes inﬁnite as x approaches a” or “ f x increases without bound as x approaches a ” This deﬁnition is illustrated graphically in Figure 12. is often read as 5E02(pp 074085) 78 ❙❙❙❙ 1/17/06 1:17 PM Page 78 CHAPTER 2 LIMITS AND RATES OF CHANGE y A similar sort of limit, for functions that become large negative as x gets close to a, is
deﬁned in Deﬁnition 5 and is illustrated in Figure 13. x=a 5 a
0 Definition Let f be deﬁned on both sides of a, except possibly at a itself. Then x lim f x y=ƒ xla means that the values of f x can be made arbitrarily large negative by taking x
sufﬁciently close to a, but not equal to a. FIGURE 13 lim ƒ=_`
xa The symbol lim x l a f x
can be read as “the limit of f x , as x approaches a,
is negative inﬁnity” or “ f x decreases without bound as x approaches a.” As an example
we have
1
lim
xl0
x2
Similar deﬁnitions can be given for the onesided inﬁnite limits
lim f x lim f x x la x la lim f x lim f x x la x la remembering that “x l a ” means that we consider only values of x that are less than a,
and similarly “x l a ” means that we consider only x a. Illustrations of these four
cases are given in Figure 14.
y y a 0 (a) lim ƒ=`
x a_ x y a 0 x (b) lim ƒ=`
x y a 0 (c) lim ƒ=_` a+ x a 0 x x (d) lim ƒ=_` a_ x a+ FIGURE 14
6 Definition The line x
a is called a vertical asymptote of the curve y
if at least one of the following statements is true: lim f x xla lim f x xla lim f x x la lim f x x la fx lim f x x la lim f x x la For instance, the yaxis is a vertical asymptote of the curve y 1 x 2 because
lim x l 0 1 x 2
. In Figure 14 the line x a is a vertical asymptote in each of the four
cases shown. In general, knowledge of vertical asymptotes is very useful in sketching
graphs. 5E02(pp 074085) 1/17/06 1:18 PM Page 79 SECTION 2.2 THE LIMIT OF A FUNCTION E XAMPLE 9 Find lim
x l3 2x
x 3 2x and lim x x l3 3 ❙❙❙❙ 79 . 3 is a small posiSOLUTION If x is close to 3 but larger than 3, then the denominator x
tive number and 2 x is close to 6. So the quotient 2 x x 3 is a large positive number.
Thus, intuitively we see that
2x
lim
x l3 x
3 y
2x y= x3
5 Likewise, if x is close to 3 but smaller than 3, then x 3 is a small negative number but
2 x is still a positive number (close to 6). So 2 x x 3 is a numerically large negative
number. Thus
2x
lim
x l3 x
3 x 0 x=3 The graph of the curve y
cal asymptote. FIGURE 15 2x x 3 is given in Figure 15. The line x EXAMPLE 10 Find the vertical asymptotes of f x
y 1
0 π
2 π 3π
2 sin x
cos x there are potential vertical asymptotes where cos x 0. In fact, since cos x l 0 as
xl
2 and cos x l 0 as x l
2 , whereas sin x is positive when x is near
2, we have
lim tan x
lim tan x
and x xl xl 2 2 2 is a vertical asymptote. Similar reasoning shows
This shows that the line x
2n 1 2, where n is an integer, are all vertical asymptotes of
that the lines x
fx
tan x. The graph in Figure 16 conﬁrms this. FIGURE 16 y=tan x  2.2 tan x. SOLUTION Because tan x 3π
π
_ 2 _π _
2 3 is a verti Exercises 1. Explain in your own words what is meant by the equation 4. For the function f whose graph is given, state the value of the lim f x given quantity, if it exists. If it does not exist, explain why.
(a) lim f x
(b) lim f x xl2 5 xl0 Is it possible for this statement to be true and yet f 2
Explain. 3? 2. Explain what it means to say that lim f x xl1 3 and xl3 (c) lim f x (d) lim f x xl3 xl3 (e) f 3
y lim f x xl1 7 In this situation is it possible that lim x l 1 f x exists? Explain. 4
2 3. Explain the meaning of each of the following. (a) lim f x
xl 3 (b) lim f x
xl4 0 2 4 x 5E02(pp 074085) 80 ❙❙❙❙ 1/17/06 1:19 PM Page 80 CHAPTER 2 LIMITS AND RATES OF CHANGE 5. Use the given graph of f to state the value of each quantity, if it exists. If it does not exist, explain why.
(a) lim f x
(b) lim f x
(c) lim f x
xl1 xl1 (d) lim f x (a) lim R x (b) lim R x (c) lim R x (d) x l2 xl1 xl5 xl 3 (e) f 5 xl5 8. For the function R whose graph is shown, state the following. lim R x xl 3 (e) The equations of the vertical asymptotes.
y y
4
2 0 2 2 x 5 x 4 6. For the function t whose graph is given, state the value of each xl 2 (d) lim f x (f) lim t x (g) lim t x (h) t 2
(k) t 0 (c) lim f x (e) lim f x xl 3 (i) lim t x ( j) lim t x (b) lim f x xl 7 xl 2 (e) lim t x 9. For the function f whose graph is shown, state the following. (a) lim f x quantity, if it exists. If it does not exist, explain why.
(a) lim t x
(b) lim t x
(c) lim t x
xl 2 0 _3 xl0 (l) lim t x (d) t 2 xl2 x l2 xl4 xl6 xl6 (f) The equations of the vertical asymptotes. xl2 y xl4 xl0 y
_7 0 _3 6 x 2
1 10. A patient receives a 150mg injection of a drug every 4 hours.
_3 _2 _1 0 1 2 3 4 x _1 The graph shows the amount f t of the drug in the bloodstream after t hours. (Later we will be able to compute the
dosage and time interval to ensure that the concentration of the
drug does not reach a harmful level.) Find
and lim f t t l 12 7. For the function t whose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.
(a) lim t t
(b) lim t t
(c) lim t t
tl0 tl0 (d) lim t t (e) lim t t (g) t 2 and explain the signiﬁcance of these onesided limits. (h) lim t t tl2 lim f t t l 12 tl0 f(t) ( f ) lim t t tl2 tl2 300 tl4 y 150 4
0 2 2 4 t 4 8 12 16 t 1 1 2 1 x to state the
; 11. Use the graph of the function f x
value of each limit, if it exists. If it does not exist, explain why.
(a) lim f x
xl0 (b) lim f x
xl0 (c) lim f x
xl0 5E02(pp 074085) 1/17/06 1:21 PM Page 81 ❙❙❙❙ S ECTION 2.2 THE LIMIT OF A FUNCTION 12. Sketch the graph of the following function and use it to deter 29. mine the values of a for which lim x l a f x exists: ■ 2
x
x fx x
1 if x
1
if 1 x
if x 1 2 1 4, 3, f3 lim f x f 2 14. lim f x 1, lim f x 1, xl0
xl2 ■ ■ ■ 2, xl3 lim f x ; 2, xl 2 1, xl0 f2 ■ 1, ■ lim f x xl2 ■ ■ ; ■ ■ ■ ; 2x x2 2x x2
0.999, , x 0, 0.5, 0.9, 0.95,
2
1.5, 1.1, 1.01, 1.001 x
2, 1 0.99, x2 ■ 1, x 0.5, 0.2, 0.1, 0.05, 0.01 36. (a) Evaluate h x xl1 ■ ■ 2 x 1x to ﬁve
1 x 1x . ■ ■ ■ ■ ■ ■ 4
x 2 20. lim xl0 1
1
■ 22. lim ■ ■ ■ ■ ■ ■ ; 5x
■ 0.04, 0.02, 0.01, 0.005, 0.003, and
x x 3 for x tan x 1, 0.5, 0.1, 0.05, tan x x
.
x3
(c) Evaluate h x for successively smaller values of x until you
ﬁnally reach 0 values for h x . Are you still conﬁdent that
your guess in part (b) is correct? Explain why you eventually obtained 0 values. (In Section 7.7 a method for evaluating the limit will be explained.)
(d) Graph the function h in the viewing rectangle 1, 1
by 0, 1 . Then zoom in toward the point where the graph
crosses the yaxis to estimate the limit of h x as x
approaches 0. Continue to zoom in until you observe distortions in the graph of h. Compare with the results of part (c).
xl0 ■ x xl0 ■ 2x
1000 (b) Guess the value of lim tan 3 x
tan 5 x
9x 1 x. 0.01, and 0.005.  Use a table of values to estimate the value of the limit.
If you have a graphing device, use it to conﬁrm your result
graphically. x6
x10 x
x decimal places.
(b) Illustrate part (a) by graphing the function y (b) Evaluate f x for x
0.001. Guess again. 19–22 21. lim ■ (b) Conﬁrm your answer to part (a) by graphing the function. lim x 2 sx 4
, x 17, 16.5, 16.1, 16.05, 16.01,
x 16
15, 15.5, 15.9, 15.95, 15.99 sx ■ x2
2 x 1000 for x 1,
0.8, 0.6, 0.4, 0.2, 0.1, and 0.05, and guess the value of x l 16 xl0 ■ 35. (a) Evaluate the function f x 18. lim 19. lim ■ 1
1
and lim 3
x l1 x
x3 1
1
(a) by evaluating f x
1 x 3 1 for values of x that
approach 1 from the left and from the right,
(b) by reasoning as in Example 9, and
(c) from a graph of f . xl0 sin x
17. lim
,
xl0 x
tan x ■ ■ function y 2 x at the point 0, 1 is lim x l 0 2 x
Estimate the slope to three decimal places. , x 2.5, 2.1, 2.05, 2.01, 2.005, 2.001,
x2 x 2
1.9, 1.95, 1.99, 1.995, 1.999 ■ ■ 34. The slope of the tangent line to the graph of the exponential x2 x l2 xl ■ x1
x sin x 33. (a) Estimate the value of the limit lim x l 0 1 ■  Guess the value of the limit (if it exists) by evaluating the
function at the given numbers (correct to six decimal places). 16. lim ■ y 15–18 15. lim ■ 0 f 0 is undeﬁned ■ ■ xl1 32. (a) Find the vertical asymptotes of the function 1
lim f x ■ 30. lim sec x 2 x l1  Sketch the graph of an example of a function f that
satisﬁes all of the given conditions. xl3 lim 31. Determine lim 13–14 13. lim f x xl 81 ■ ■ ■ ; 37. Graph the function f x
23–30  23. lim 6 xl5 x 25. lim 2
x 24. lim 5
x
12 x l1 27. lim xl 2 x of Example 4 in the viewsin
ing rectangle 1, 1 by 1, 1 . Then zoom in toward the
origin several times. Comment on the behavior of this function. Determine the inﬁnite limit. x
x2 x xl5 26. lim xl0 1
2 6
x 5 x
x2 x 28. lim csc x
xl 1
2 38. In the theory of relativity, the mass of a particle with velocity
v is m m0 s1
v2 c2 where m 0 is the rest mass of the particle and c is the speed of
light. What happens as v l c ? 5E02(pp 074085) 82 ❙❙❙❙ 1/17/06 1:21 PM Page 82 CHAPTER 2 LIMITS AND RATES OF CHANGE ; 39. Use a graph to estimate the equations of all the vertical asymp ; 40. (a) Use numerical and graphical evidence to guess the value of
the limit totes of the curve lim y tan 2 sin x xl1 x 1
1 (b) How close to 1 does x have to be to ensure that the function
in part (a) is within a distance 0.5 of its limit? Then ﬁnd the exact equations of these asymptotes.  2.3 x3
sx Calculating Limits Using the Limit Laws
In Section 2.2 we used calculators and graphs to guess the values of limits, but we saw that
such methods don’t always lead to the correct answer. In this section we use the following
properties of limits, called the Limit Laws, to calculate limits.
Limit Laws Suppose that c is a constant and the limits lim f x and xla lim t x xla exist. Then
1. lim f x tx 2. lim f x tx xla xla 3. lim c f x
xla lim f x xla lim f x xla xla lim f x xla xla fx
tx lim t x xla c lim f x 4. lim f x t x 5. lim lim t x xla xla lim t x xla lim f x xla lim t x if lim t x
xla 0 xla These ﬁve laws can be stated verbally as follows:
Sum Law 1. The limit of a sum is the sum of the limits. Difference Law 2. The limit of a difference is the difference of the limits. Constant Multiple Law 3. The limit of a constant times a function is the constant times the limit of the function.
Product Law 4. The limit of a product is the product of the limits. Quotient Law 5. The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0).
It is easy to believe that these properties are true. For instance, if f x is close to L and
t x is close to M, it is reasonable to conclude that f x
t x is close to L M. This gives
us an intuitive basis for believing that Law 1 is true. In Section 2.4 we give a precise definition of a limit and use it to prove this law. The proofs of the remaining laws are given
in Appendix F. 5E02(pp 074085) 1/17/06 1:22 PM Page 83 S ECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS y following limits, if they exist. 1 (a) lim
1 83 EXAMPLE 1 Use the Limit Laws and the graphs of f and t in Figure 1 to evaluate the f 0 ❙❙❙❙ x xl 2 fx 5t x (b) lim f x t x (c) lim xl1 xl2 fx
tx SOLUTION g (a) From the graphs of f and t we see that
lim f x 1 xl 2 FIGURE 1 and lim t x 1 xl 2 Therefore, we have
lim xl 2 fx 5t x lim f x lim 5t x lim f x (b) We see that lim x l 1 f x
right limits are different: 5 (by Law 3) xl 2 xl 2 1 (by Law 1) 5 lim t x xl 2 xl 2 1 4 2. But lim x l 1 t x does not exist because the left and lim t x 2 xl1 lim t x xl1 1 So we can’t use Law 4. The given limit does not exist, since the left limit is not equal to
the right limit.
(c) The graphs show that
lim f x xl2 1.4 and lim t x xl2 0 Because the limit of the denominator is 0, we can’t use Law 5. The given limit does not
exist because the denominator approaches 0 while the numerator approaches a nonzero
number.
If we use the Product Law repeatedly with t x Power Law n 6. lim f x
x la n [ lim f x ]
x la f x , we obtain the following law. where n is a positive integer In applying these six limit laws, we need to use two special limits:
7. lim c
xla 8. lim x c xla a These limits are obvious from an intuitive point of view (state them in words or draw
graphs of y c and y x), but proofs based on the precise deﬁnition are requested in the
exercises for Section 2.4.
If we now put f x
x in Law 6 and use Law 8, we get another useful special limit.
9. lim x n
xla an where n is a positive integer 5E02(pp 074085) 84 ❙❙❙❙ 1/17/06 1:23 PM Page 84 CHAPTER 2 LIMITS AND RATES OF CHANGE A similar limit holds for roots as follows. (For square roots the proof is outlined in Exercise 37 in Section 2.4.) n
10. lim sx n
sa xla where n is a positive integer (If n is even, we assume that a 0.) More generally, we have the following law, which is proved as a consequence of Law 10
in Section 2.5. n
11. lim sf x) Root Law x la n s lim f
x la x) where n is a positive integer [If n is even, we assume that lim f x 0. x la Explore limits like these interactively.
Resources / Module 2
/ The Essential Examples
/ Examples D and E EXAMPLE 2 Evaluate the following limits and justify each step. (a) lim 2 x 2
x l5 3x 4 (b) lim x3 xl 2 2x2 1
5 3x SOLUTION lim 2 x 2 (a) x l5 3x 4 lim 2 x 2 lim 3x x l5 2 lim x 2 3 lim x x l5 2 52 lim 4 x l5 lim 4 x l5 35 (by Laws 2 and 1) x l5 (by 3) x l5 4 (by 9, 8, and 7) 39
(b) We start by using Law 5, but its use is fully justiﬁed only at the ﬁnal stage when we
see that the limits of the numerator and denominator exist and the limit of the denominator is not 0.
lim xl 2 x3 2x 2 1
5 3x lim x 3 2x2 xl 2 lim 5 2 lim x 2 xl 2 lim 5 3 5 lim 1 xl 2 xl 2 2 (by Law 5) 3x xl 2 lim x 3 1 xl 2 3 lim x 2 22
32 (by 1, 2, and 3) xl 2 1 (by 9, 8, and 7) 1
11
If we let f x
2 x 2 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 part (b). The functions in Example 2 are a polynomial and a rational function, respectively, and similar use of the Limit Laws proves that
NOTE ■ 5E02(pp 074085) 1/17/06 1:23 PM Page 85 S ECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS  NEWTON AND LIMITS
Isaac Newton was born on Christmas Day in
1642, the year of Galileo’s death. When he
entered Cambridge University in 1661 Newton
didn’t know much mathematics, but he learned
quickly by reading Euclid and Descartes and
by attending the lectures of Isaac Barrow.
Cambridge was closed because of the plague in
1665 and 1666, and Newton returned home to
reﬂect on what he had learned. Those two years
were amazingly productive for at that time he
made four of his major discoveries: (1) his
representation of functions as sums of inﬁnite
series, including the binomial theorem; (2) his
work on differential and integral calculus; (3) his
laws of motion and law of universal gravitation;
and (4) his prism experiments on the nature of
light and color. Because of a fear of controversy
and criticism, he was reluctant to publish his discoveries and it wasn’t until 1687, at the urging of
the astronomer Halley, that Newton published
Principia Mathematica. In this work, the greatest
scientiﬁc treatise ever written, Newton set forth
his version of calculus and used it to investigate
mechanics, ﬂuid dynamics, and wave motion,
and to explain the motion of planets and comets.
The beginnings of calculus are found in the
calculations of areas and volumes by ancient
Greek scholars such as Eudoxus and Archimedes.
Although aspects of the idea of a limit are
implicit in their “method of exhaustion,” Eudoxus
and Archimedes never explicitly formulated the
concept of a limit. Likewise, mathematicians
such as Cavalieri, Fermat, and Barrow, the immediate precursors of Newton in the development
of calculus, did not actually use limits. It was
Isaac Newton who was the ﬁrst to talk explicitly
about limits. He explained that the main idea
behind limits is that quantities “approach nearer
than by any given difference.” Newton stated
that the limit was the basic concept in calculus,
but it was left to later mathematicians like
Cauchy to clarify his ideas about limits. ❙❙❙❙ 85 direct substitution always works for such functions (see Exercises 53 and 54). We state this
fact as follows.
Direct Substitution Property If f is a polynomial or a rational function and a is in the domain of f , then
lim f x fa xla Functions with the Direct Substitution Property are called continuous at a and will be
studied in Section 2.5. However, not all limits can be evaluated by direct substitution, as
the following examples show.
EXAMPLE 3 Find lim xl1 x2
x 1
.
1 x 2 1 x 1 . We can’t ﬁnd the limit by substituting x 1
because f 1 isn’t deﬁned. Nor can we apply the Quotient Law because the limit of the
denominator is 0. Instead, we need to do some preliminary algebra. We factor the numerator as a difference of squares:
SOLUTION Let f x x2
x 1
1 x 1x
x1 1 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. Therefore, we can cancel the common factor and compute the limit as follows:
lim xl1 x2
x 1
1 lim x xl1 1x
x1 lim x 1 1 1 2 xl1 1 The limit in this example arose in Section 2.1 when we were trying to ﬁnd the tangent to
the parabola y x 2 at the point 1, 1 .
NOTE In Example 3 we were able to compute the limit by replacing the given function f x
x 2 1 x 1 by a simpler function, t x
x 1, with the same limit.
This is valid because f x
t 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. In general,
if f x
t x when x a, then
■ lim f x xla lim t x xla EXAMPLE 4 Find lim t x where
x l1 tx x 1 if x
if x 1
1 S OLUTION Here t is deﬁned at x
1 and t 1
, but the value of a limit as x
approaches 1 does not depend on the value of the function at 1. Since t x
x 1 for 5E02(pp 086097) 86 ❙❙❙❙ 1/17/06 12:48 PM Page 86 CHAPTER 2 LIMITS AND RATES OF CHANGE x 1, we have
lim t x lim x xl1 x 1 xl1 2 Note that the values of the functions in Examples 3 and 4 are identical except when
1 (see Figure 2) and so they have the same limit as x approaches 1.
y y y=ƒ 3
2 The graphs of the functions f (from
Example 3) and g (from Example 4) 2 1 FIGURE 2 y=© 3 1 0 1 EXAMPLE 5 Evaluate lim 2 h2
h 3 hl0 0 x 3 9 1 2 x 3 . SOLUTION If we deﬁne h2
h 3 Fh 9 then, as in Example 3, we can’t compute lim h l 0 F h by letting h
undeﬁned. But if we simplify F h algebraically, we ﬁnd that
Fh 9 lim st 2 tl0 h2 6h 9
t2 6 h h 0 when letting h approach 0.) Thus
h2
h 3 hl0 EXAMPLE 6 Find lim 9 h (Recall that we consider only h Explore a limit like this one interactively.
Resources / Module 2
/ The Essential Examples
/ Example C h2 6h 0 since F 0 is 3 9 lim 6 h hl0 6 . SOLUTION 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: lim
tl0 st 2 9
t 2 3 lim st 2
t tl0 lim
tl0 lim
tl0 9 3 2 t2 9
t (st 2 9 9
3) 2 st 2 1
9 st 2
st 2 9
9 3
3
t2 lim t (st
2 tl0 2 9 3) 1
3 s lim
tl0 t 2 1
9 3 3 3 This calculation conﬁrms the guess that we made in Example 2 in Section 2.2. 1
6 5E02(pp 086097) 1/17/06 12:48 PM Page 87 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ❙❙❙❙ 87 Some limits are best calculated by ﬁrst ﬁnding the left and righthand limits. The following theorem is a reminder of what we discovered in Section 2.2. It says that a twosided
limit exists if and only if both of the onesided limits exist and are equal. 1 Theorem lim f x L xla if and only if lim f x xla L lim f x xla When computing onesided limits, we use the fact that the Limit Laws also hold for
onesided limits.
EXAMPLE 7 Show that lim x 0. xl0 SOLUTION Recall that x x
Since x  The result of Example 7 looks plausible
from Figure 3. x for x if x
x if x lim x 0 0, we have
lim x y xl0 y= x For x 0 we have x xl0 x and so
lim x lim xl0 0 x x xl0 lim x 0 xl0 E XAMPLE 8 Prove that lim xl0 x
does not exist.
x lim x
x xl0 lim SOLUTION y 0 Therefore, by Theorem 1, FIGURE 3 x
y= x 0
0 x
x xl0 xl0 lim x
x lim 1 1 lim 1 xl0 1
0 x xl0 lim x
x xl0 1 _1 FIGURE 4 Since the right and lefthand limits are different, it follows from Theorem 1 that
lim x l 0 x x does not exist. The graph of the function f x
x x is shown in
Figure 4 and supports the onesided limits that we found.
EXAMPLE 9 If if x
if x sx 4
8 2x fx 4
4 determine whether lim x l 4 f x exists.
 It is shown in Example 3 in
Section 2.4 that lim x l 0 sx 0. SOLUTION Since f x sx 4 for x lim f x xl4 4, we have
lim s x xl4 4 s4 4 0 5E02(pp 086097) ❙❙❙❙ 88 1/17/06 12:48 PM Page 88 CHAPTER 2 LIMITS AND RATES OF CHANGE Since f x 8 2 x for x y 4, we have lim f x lim 8 xl4 2x xl4 8 24 0 The right and lefthand limits are equal. Thus, the limit exists and
0 x 4 lim f x 0 xl4 The graph of f is shown in Figure 5. FIGURE 5 EXAMPLE 10 The greatest integer function is deﬁned by x that is less than or equal to x. (For instance, 4
4, 4.8
1
1.) Show that lim x l 3 x does not exist.
2
 Other notations for x are x and ⎣ x⎦ . SOLUTION The graph of the greatest integer function is shown in Figure 6. Since x for 3 y x lim x x l3 3 y=[ x] Since x 3 4, we have 4 2 the largest integer
4,
3, s2
1, 2 for 2 x lim 3 3 lim 2 2 x l3 3, we have 1
0 1 2 3 4 5 lim x x x l3 x l3 Because these onesided limits are not equal, lim x l 3 x does not exist by Theorem 1.
FIGURE 6 The next two theorems give two additional properties of limits. Their proofs can be
found in Appendix F. Greatest integer function Theorem If f x
t x when x is near a (except possibly at a) and the limits
of f and t both exist as x approaches a, then
2 lim f x lim t x xla 3 The Squeeze Theorem If f x tx xla h x when x is near a (except possibly at a) and
lim f x y xla h
g
L f
0 FIGURE 7 a x then lim h x xla lim t x xla L L The Squeeze Theorem, which is sometimes called the Sandwich Theorem or the
Pinching Theorem, is illustrated by Figure 7. It says that if t x is squeezed between f x
and h x near a, and if f and h have the same limit L at a, then t is forced to have the same
limit L at a. 5E02(pp 086097) 1/17/06 12:49 PM Page 89 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS E XAMPLE 11 Show that lim x 2 sin
xl0 1
x ❙❙❙❙ 0. SOLUTION First note that we cannot use lim x 2 sin xl0 1
x lim x 2 lim sin xl0 xl0 1
x because lim x l 0 sin 1 x does not exist (see Example 4 in Section 2.2). However, since
1 sin 1
x 1 we have, as illustrated by Figure 8,
x2 x 2 sin 1
x x2 y y=≈ 1 y=≈ sin x Watch an animation of a similar limit.
Resources / Module 2
/ Basics of Limits
/ Sound of a Limit that Exists x 0 y=_≈ FIGURE 8 We know that
lim x 2 xl0 x 2, t x Taking f x
obtain 0 and x 2 sin 1 x , and h x
lim x 2 sin xl0  2.3 1
x x2 0 x 2 in the Squeeze Theorem, we 0 Exercises 1. Given that lim f x
x la 3
(c) lim sh x 3 lim t x
x la 0 lim h x
x la 8 ﬁnd the limits that exist. If the limit does not exist, explain
why.
(a) lim f x
x la lim xl0 hx (b) lim f x
x la 2 xla f
h
f
(g) lim
x la t
(e) lim
x la x
x
x
x 1
fx
tx
(f ) lim
x la f x
2f x
(h) lim
x la h x
fx
(d) lim
x la 89 5E02(pp 086097) 90 ❙❙❙❙ 1/17/06 12:50 PM Page 90 CHAPTER 2 LIMITS AND RATES OF CHANGE 2. The graphs of f and t are given. Use them to evaluate each 21. lim limit, if it exists. If the limit does not exist, explain why.
y tl9 y 23. lim y=ƒ x 1 (a) lim f x 1 0 x l0 xl 1 29. lim
tl0 (f ) lim s3 x l2 3–9 fx x l1 7 1
t s1 ■ 2x2 xl 2 1 4. lim 5x 1 4 x3 7. lim 1 3x
4 x 2 3x 4 1 x l1 x l2 x2 6x
1 ul 2 ■ ■ ■ ■ ■ ■ ■ ■ 3 t 3 t ■ ■ ■ ■ ■ 10. (a) What is wrong with the following equation? x2 x
x 6
2 x ■ ■ x2 x l2 x
x 6
2 5 s1 x
3x 11. lim x 2 13. lim s3 ■ lim x tl 3 x
x xl 4 6 2 x xl4 3 16. lim 5x
3x x2 14. lim 9
7t x
x2 12. lim 2
t 2t 2 2 x xl 1 x2 4
4 4x
3x 4 2
2 17. lim 4 hl0 19. lim hl0 1 h
h 16 h4
h 1 3 18. lim
x l1 20. lim h l0 s3 x
x x 2 sin 4x
3x 4 x
x2 1
1 2 h3
h 0 x Illustrate by graphing the functions f, t, and h (in the notation
of the Squeeze Theorem) on the same screen.
35. If 1 x2 fx 36. If 3x fx 2x x3 37. Prove that lim x 4 cos
x l0 38. Prove that lim sx 1
x l0 2 ■ 1 ; 34. Use the Squeeze Theorem to show that 3 x l2 x l0 6 ■ to estimate the value of lim x l 0 f x to two decimal places.
(b) Use a table of values of f x to estimate the limit to four
decimal places.
(c) Use the Limit Laws to ﬁnd the exact value of the limit. lim sx 3 2 x2 x l2 15. lim x
x x l2 ■ lim x l 0 x 2 cos 20 x 0. Illustrate by graphing the functions
fx
x 2, t x
x 2 cos 20 x, and h x
x 2 on the same
screen. Evaluate the limit, if it exists.  ■ ; 32. (a) Use a graph of is correct.
11–30 ■ ; 33. Use the Squeeze Theorem to show that 3 (b) In view of part (a), explain why the equation
lim 1 3 by graphing the function f x
x (s1 3x 1).
(b) Make a table of values of f x for x close to 0 and guess
the value of the limit.
(c) Use the Limit Laws to prove that your guess is correct. 6 ■ 1 h sx x 2
1 sx x2 xl4 3 t h x l1 fx
9. lim s16 1 30. lim 1
t x l0 3 8. lim su 4 t2 hl0 lim 4 3u tl 1 1 ; 31. (a) Estimate the value of 1 6. lim t 2 5. lim x 2
xl3 2x x 1
t 28. lim Evaluate the limit and justify each step by indicating the
appropriate Limit Law(s).
3. lim 3x 4 16
2 tl0  2 x4
x x l2 81
3 ■ (e) lim x 3f x 3 x2
sx x l9 fx
tx (d) lim 24. lim 26. lim 2
x h
h h l0 1
x
x 27. lim tx x l1 (c) lim f x t x sx s1 22. lim 1
4
25. lim
xl 4 4 x 1 (b) lim f x tx x l2 t
st x l7 y=© 1 9
3 2 for all x, ﬁnd lim x l 2 for 0
2
x 1 f x. 2, evaluate lim x l 1 f x . x 0.
sin2 2 x 0. 39–44  Find the limit, if it exists. If the limit does not exist,
explain why. 8 39. lim x
xl 4 4 40. lim xl 4 x
x 4
4 5E02(pp 086097) 1/17/06 12:51 PM Page 91 ❙❙❙❙ S ECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS x
x 41. lim
x l2 43. lim
x l0 ■ 2
2 1
x ■ 42. lim x l 1.5 1
x
■ 2x 2
2x x l0 ■ ■ ■ ■ ■ (b) If n is an integer, evaluate
(i) lim f x
(ii) lim f x 3x
3 1
x 44. lim x ln 51. If f x ■ ■ ■ 45. The signum (or sign) function, denoted by sgn, is deﬁned by 1
0
1 sgn x if x
if x
if x xl0 xl0 x , show that lim x l 2 f x exists but is not 52. In the theory of relativity, the Lorentz contraction formula xl0 x2
1 fx x2
x 54. If r is a rational function, use Exercise 53 to show that if x
if x r a for every number a in the domain of r. x l1 x2
0 fx 2
2 prove that lim x l 0 f x 1
.
1 57. Show by means of an example that limx l a f x t x may exist even though neither lim x l a f x nor limx l a t x exists.
58. Evaluate lim (ii) lim F x x l2 s6
s3 x
x 2
.
1 x l1 59. Is there a number a such that lim xl 2 48. Let x
x2
8 if x
if 0
x if x 0
x
2 xl0 (v) lim h x xl2 xl2 x l1 (vi) lim h x
x l2 y 49. (a) If the symbol denotes the greatest integer function
deﬁned in Example 10, evaluate
(i) lim x
(ii) lim x
(iii) lim x
xl 2 (b) If n is an integer, evaluate
(i) lim x
(ii) lim x
x ln xln (c) For what values of a does lim x l a x exist?
x
x.
(a) Sketch the graph of f. ax a 3
x2 x 1 2 y 2 1 and a shrinking circle C2 with radius r and
center the origin. P is the point 0, r , Q is the upper point of
intersection of the two circles, and R is the point of intersection
of the line PQ and the xaxis. What happens to R as C2 shrinks,
that is, as r l 0 ? (b) Sketch the graph of h. xl 2 x2 60. The ﬁgure shows a ﬁxed circle C1 with equation 2 xl0 (iv) lim h x 3x 2 exists? If so, ﬁnd the value of a and the value of the limit. (a) Evaluate each of the following limits, if it exists.
(i) lim h x
(ii) lim h x
(iii) lim h x 50. Let f x 0. t x may
exist even though neither limx l a f x nor limx l a t x exists. (b) Does lim x l 1 F x exist?
(c) Sketch the graph of F . hx if x is rational
if x is irrational 56. Show by means of an example that lim x l a f x (a) Find
(i) lim F x pa. 55. If (a) Find lim x l 2 f x and lim x l 2 f x .
(b) Does lim x l 2 f x exist?
(c) Sketch the graph of f .
47. Let F x v2 c2 53. If p is a polynomial, show that lim xl a p x 46. Let 4
x L 0 s1 expresses the length L of an object as a function of its velocity
v with respect to an observer, where L 0 is the length of the
object at rest and c is the speed of light. Find lim v l c L and
interpret the result. Why is a lefthand limit necessary? lim x l a r x (iv) lim sgn x xl0 x
equal to f 2 . L 0
0
0 (a) Sketch the graph of this function.
(b) Find each of the following limits or explain why it does not
exist.
(i) lim sgn x
(ii) lim sgn x
(iii) lim sgn x x ln (c) For what values of a does lim x l a f x exist? 1
x ■ 91 x l 2.4 P Q C™ 0 R
C¡ x 5E02(pp 086097) 92 ❙❙❙❙ 1/17/06 12:52 PM Page 92 CHAPTER 2 LIMITS AND RATES OF CHANGE  2.4 The Precise Definition of a Limit
The intuitive deﬁnition of a limit given in Section 2.2 is inadequate for some purposes
because such phrases as “x is close to 2” and “ f x gets closer and closer to L” are vague.
In order to be able to prove conclusively that
cos 5 x
10,000 lim x 3 xl0 0.0001 or lim xl0 sin x
x 1 we must make the deﬁnition of a limit precise.
To motivate the precise deﬁnition of a limit, let’s consider the function
2x
6 fx 1 if x
if x 3
3 Intuitively, it is clear that when x is close to 3 but x 3, then f x is close to 5, and so
lim x l 3 f x
5.
To obtain more detailed information about how f x varies when x is close to 3, we ask
the following question:
How close to 3 does x have to be so that f x differs from 5 by less than 0.l?  It is traditional to use the Greek letter
(delta) in this situation. The distance from x to 3 is x 3 and the distance from f x to 5 is f x
problem is to ﬁnd a number such that
fx 5 If x 3
0, then x
ber such that 0.1 x that is, 5 3 fx x 3 but x 3 3, so an equivalent formulation of our problem is to ﬁnd a numfx Notice that if 0 if 5 , so our 0.1 0.1 2 5 2x fx 5 1
0.1 if 0 x 3 0.05, then
5
if 2x 6 2x 0 x 3 3 0.1 0.05 Thus, an answer to the problem is given by
0.05; that is, if x is within a distance of
0.05 from 3, then f x will be within a distance of 0.1 from 5.
If we change the number 0.l in our problem to the smaller number 0.01, then by using
the same method we ﬁnd that f x will differ from 5 by less than 0.01 provided that x differs from 3 by less than (0.01) 2 0.005:
fx 5 0.01 if 0 x 3 0.005 0.001 if 0 x 3 0.0005 Similarly,
fx 5 The numbers 0.1, 0.01, and 0.001 that we have considered are error tolerances that we
might allow. For 5 to be the precise limit of f x as x approaches 3, we must not only be
able to bring the difference between f x and 5 below each of these three numbers; we 5E02(pp 086097) 1/17/06 12:52 PM Page 93 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT y ƒ
is in
here 93 must be able to bring it below any positive number. And, by the same reasoning, we can!
If we write (the Greek letter epsilon) for an arbitrary positive number, then we ﬁnd as
before that 5+∑
5 5∑ fx 1 0 x 3 3∂ 3+∂ when x is in here
(x≠3)
FIGURE 1 ❙❙❙❙ 5 if 0 x 3 2 This is a precise way of saying that f x is close to 5 when x is close to 3 because (1) says
that we can make the values of f x within an arbitrary distance from 5 by taking the values of x within a distance 2 from 3 (but x 3).
Note that (1) can be rewritten as
5 fx 5 whenever 3 x 3 x 3 and this is illustrated in Figure 1. By taking the values of x ( 3) to lie in the interval
3
,3
we can make the values of f x lie in the interval 5
,5
.
Using (1) as a model, we give a precise deﬁnition of a limit.
2 Definition Let f be a function deﬁned on some open interval that contains the
number a, except possibly at a itself. Then we say that the limit of f x as x
approaches a is L, and we write lim f x xla if for every number 0 there is a number fx L whenever L
0 such that
0 x a Another way of writing the last line of this deﬁnition is
if 0 x a then fx L Since x a is the distance from x to a and f x
L is the distance from f x to L,
and since can be arbitrarily small, the deﬁnition of a limit can be expressed in words
as follows:
L means that the distance between f x and L can be made arbitrarily small
lim x l a f x
by taking the distance from x to a sufﬁciently small (but not 0). Alternatively,
L means that the values of f x can be made as close as we please to L
lim x l a f x
by taking x close enough to a (but not equal to a). We can also reformulate Deﬁnition 2 in terms of intervals by observing that the inequality x a
is equivalent to
xa
, which in turn can be written
as a
xa
. Also 0
x a is true if and only if x a 0, that is,
x a. Similarly, the inequality f x
L
is equivalent to the pair of inequalities
L
fx
L
. Therefore, in terms of intervals, Deﬁnition 2 can be stated
as follows:
L means that for every
lim x l a f x
0 (no matter how small
and x
0 such that if x lies in the open interval a
,a
the open interval L
,L
. is) we can ﬁnd
a, then f x lies in 5E02(pp 086097) ❙❙❙❙ 94 1/17/06 12:53 PM Page 94 CHAPTER 2 LIMITS AND RATES OF CHANGE We interpret this statement geometrically by representing a function by an arrow diagram
as in Figure 2, where f maps a subset of onto another subset of . f
FIGURE 2 x a f(a) ƒ The deﬁnition of limit says that if any small interval L
,L
is given around L,
then we can ﬁnd an interval a
,a
around a such that f maps all the points in
a
,a
(except possibly a) into the interval L
,L
. (See Figure 3.)
f
x
FIGURE 3 ƒ
a a∂ a+∂ L∑ L L+∑ Another geometric interpretation of limits can be given in terms of the graph of a function. If
0 is given, then we draw the horizontal lines y L
and y L
and
the graph of f (see Figure 4). If lim x l a f x
L, then we can ﬁnd a number
0 such
that if we restrict x to lie in the interval a
and take x a, then the curve
,a
lies between the lines y L
and y L
. (See Figure 5.) You can see that
y fx
if such a has been found, then any smaller will also work.
It is important to realize that the process illustrated in Figures 4 and 5 must work for
every positive number no matter how small it is chosen. Figure 6 shows that if a smaller
is chosen, then a smaller may be required.
y y y y=ƒ L+∑
y=L+∑ L ƒ
is in
here ∑
∑ y=L∑ y=L+∑
L y=L+∑ ∑
∑ y=L∑ y=L∑
L∑ 0 a x 0 a∂ 0 x a a∂ a+∂ x a a+∂ when x is in here
(x≠ a)
F IGURE 4 FIGURE 5 FIGURE 6 EXAMPLE 1 Use a graph to ﬁnd a number x3 5x 6 2 0.2 such that
whenever In other words, ﬁnd a number that corresponds to
the function f x
x 3 5x 6 with a 1 and L x 1 0.2 in the deﬁnition of a limit for
2. 5E02(pp 086097) 1/17/06 12:53 PM Page 95 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT 15 ❙❙❙❙ 95 SOLUTION A graph of f is shown in Figure 7; we are interested in the region near the point 1, 2 . Notice that we can rewrite the inequality
x3
_3 3 6 2 1.8 as 5x
x3 5x 6 0.2
2.2 So we need to determine the values of x for which the curve y x 3 5x 6 lies
between the horizontal lines y 1.8 and y 2.2. Therefore, we graph the curves
y x 3 5x 6, y 1.8, and y 2.2 near the point 1, 2 in Figure 8. Then we use
the cursor to estimate that the xcoordinate of the point of intersection of the line
y 2.2 and the curve y x 3 5x 6 is about 0.911. Similarly, y x 3 5x 6
intersects the line y 1.8 when x 1.124. So, rounding to be safe, we can say that _5 FIGURE 7
2.3
y=2.2
y=˛5x+6 y=1.8
0.8
1.7 FIGURE 8 x3 1.8 (1, 2) 1.2 5x 6 2.2 whenever 0.92 x 1.12 This interval 0.92, 1.12 is not symmetric about x 1. The distance from x 1 to the
left endpoint is 1 0.92 0.08 and the distance to the right endpoint is 0.12. We can
choose to be the smaller of these numbers, that is,
0.08. Then we can rewrite our
inequalities in terms of distances as follows:
x3 5x 6 2 0.2 whenever x 1 0.08 This just says that by keeping x within 0.08 of 1, we are able to keep f x within 0.2
of 2.
Although we chose
0.08, any smaller positive value of would also have
worked.
The graphical procedure in Example 1 gives an illustration of the deﬁnition for
0.2,
but it does not prove that the limit is equal to 2. A proof has to provide a for every .
In proving limit statements it may be helpful to think of the deﬁnition of limit as a challenge. First it challenges you with a number . Then you must be able to produce a suitable . You have to be able to do this for every
0, not just a particular .
Imagine a contest between two people, A and B, and imagine yourself to be B. Person
A stipulates that the ﬁxed number L should be approximated by the values of f x to within
a degree of accuracy (say, 0.01). Person B then responds by ﬁnding a number such that
fx
L
xa
whenever 0
. Then A may become more exacting and
challenge B with a smaller value of (say, 0.0001). Again B has to respond by ﬁnding a
corresponding . Usually the smaller the value of , the smaller the corresponding value
L.
of must be. If B always wins, no matter how small A makes , then lim x l a f x
E XAMPLE 2 Prove that lim 4 x
x l3 5 7. SOLUTION
1. Preliminary analysis of the problem ( guessing a value for positive number. We want to ﬁnd a number
4x
But 4x 5 7 5 7 4x
4x 3 whenever
12 4x ). Let be a given such that 3 whenever 0
4x
0 x 3 3 . Therefore, we want
x 3 5E02(pp 086097) 96 ❙❙❙❙ 1/17/06 12:54 PM Page 96 CHAPTER 2 LIMITS AND RATES OF CHANGE that is, 3 whenever 4 0 This suggests that we should choose
4.
2. Proof (showing that this works ). Given
, then
0
x3 y y=4x5 7+∑ x x 3 0, choose 4. If 7 4x 7∑ 5 7 4x 12 4x 3 4 4 4 Thus
4x 5 7 whenever 0 x 3 Therefore, by the deﬁnition of a limit,
0 3 3∂
FIGURE 9 x lim 4 x 3+∂ x l3 5 7 This example is illustrated by Figure 9.
Note that in the solution of Example 2 there were two stages—guessing and proving.
We made a preliminary analysis that enabled us to guess a value for . But then in the second stage we had to go back and prove in a careful, logical fashion that we had made a
correct guess. This procedure is typical of much of mathematics. Sometimes it is necessary to ﬁrst make an intelligent guess about the answer to a problem and then prove that
the guess is correct.
The intuitive deﬁnitions of onesided limits that were given in Section 2.2 can be precisely reformulated as follows.
3 Definition of LeftHand Limit lim f x xla if for every number
fx 4 L 0 there is a number
L 0 such that
a whenever x a Definition of RightHand Limit lim f x xla if for every number
fx 0 there is a number
L whenever L
0 such that
a x a Notice that Deﬁnition 3 is the same as Deﬁnition 2 except that x is restricted to lie in
the left half a
. In Deﬁnition 4, x is restricted to lie
, a of the interval a
,a
in the right half a, a
of the interval a
,a
.
E XAMPLE 3 Use Deﬁnition 4 to prove that lim s x
xl0 0. 5E02(pp 086097) 1/17/06 12:54 PM Page 97 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT  CAUCHY AND LIMITS
After the invention of calculus in the 17th century, there followed a period of free development
of the subject in the 18th century. Mathematicians like the Bernoulli brothers and Euler were
eager to exploit the power of calculus and boldly
explored the consequences of this new and
wonderful mathematical theory without worrying
too much about whether their proofs were completely correct.
The 19th century, by contrast, was the Age of
Rigor in mathematics. There was a movement to
go back to the foundations of the subject—to
provide careful deﬁnitions and rigorous proofs.
At the forefront of this movement was the
French mathematician AugustinLouis Cauchy
(1789–1857), who started out as a military engineer before becoming a mathematics professor
in Paris. Cauchy took Newton’s idea of a limit,
which was kept alive in the 18th century by the
French mathematician Jean d’Alembert, and
made it more precise. His deﬁnition of a limit
reads as follows: “When the successive values
attributed to a variable approach indeﬁnitely a
ﬁxed value so as to end by differing from it by
as little as one wishes, this last is called the
limit of all the others.” But when Cauchy used
this deﬁnition in examples and proofs, he often
employed deltaepsilon inequalities similar to
the ones in this section. A typical Cauchy proof
starts with: “Designate by and two very
small numbers; . . .” He used because of the
correspondence between epsilon and the French
word erreur. Later, the German mathematician
Karl Weierstrass (1815–1897) stated the deﬁnition of a limit exactly as in our Deﬁnition 2. SOLUTION
1. Guessing a value for . Let be a given positive number. Here a
such that so we want to ﬁnd a number
sx 0 0 whenever 0 0 and L 97 0, x sx that is, whenever ❙❙❙❙ x or, squaring both sides of the inequality s x
2 x , we get whenever
. 2 0, let
s s so x 2 This suggests that we should choose
2. Showing that this works. Given
sx 0 sx x , then 2 0 According to Deﬁnition 4, this shows that lim x l 0 sx
EXAMPLE 4 Prove that lim x 2 . If 0 0. 9. xl3 SOLUTION
1. Guessing a value for . Let 0 be given. We have to ﬁnd a number 0 such that
x2
To connect x 2
want 9 whenever
3 we write x 2 9 with x
x 3 x 0 3 x 3 x 3x 9 whenever 0 x Notice that if we can ﬁnd a positive constant C such that x
x 3 x 3 Cx 3 . Then we 3 3 C, then 3 and we can make C x 3
by taking x 3
C
.
We can ﬁnd such a number C if we restrict x to lie in some interval centered at 3.
In fact, since we are interested only in values of x that are close to 3, it is reasonable
to assume that x is within a distance l from 3, that is, x 3
1. Then 2 x 4,
so 5 x 3 7. Thus, we have x 3
7, and so C 7 is a suitable choice for
the constant.
But now there are two restrictions on x 3 , namely
x 3 1 and x 3 C 7 To make sure that both of these inequalities are satisﬁed, we take to be the smaller of
the two numbers 1 and 7. The notation for this is
min 1, 7 .
2. Showing that this works. Given
0, let
min 1, 7 . If 0
x3
,
then x 3
1?2 x 4? x 3
7 (as in part l). We also have
x3
7, so
x2 9 This shows that lim x l 3 x 2 9. x 3 x 3 7 7 5E02(pp 098109) 98 ❙❙❙❙ 1/19/06 4:19 PM Page 98 CHAPTER 2 LIMITS AND RATES OF CHANGE As Example 4 shows, it is not always easy to prove that limit statements are true
using the , deﬁnition. In fact, if we had been given a more complicated function such
as f x
6 x 2 8 x 9 2 x 2 1 , a proof would require a great deal of ingenuity.
Fortunately this is unnecessary because the Limit Laws stated in Section 2.3 can be proved
using Deﬁnition 2, and then the limits of complicated functions can be found rigorously
from the Limit Laws without resorting to the deﬁnition directly.
For instance, we prove the Sum Law: If lim x l a f x
L and lim x l a t x
M both
exist, then
lim f x tx xla L M The remaining laws are proved in the exercises and in Appendix F.
Proof of the Sum Law Let fx b tx L M 0 such that whenever 0 x a Using the Triangle Inequality we can write  Triangle Inequality:
a 0 be given. We must ﬁnd a b fx 5 tx L M fx (See Appendix A.) L fx
We make f x
tx
LM
and t x
M less than 2.
Since 2 0 and lim x l a f x
fx L Let min
if 0 , 1 x 2 L whenever 0 whenever 2 M by making each of the terms f x a 1 0 such that 2 0 x a 1 and 0 tx M 2 . Notice that
a and so then 0
fx L x a and 2 x a 2 Therefore, by (5),
fx tx L M fx
2 L tx M 2 To summarize,
fx L 0 such that 1 x M , there exists a number M M tx L, there exists a number 2 Similarly, since lim x l a t x
tx less than tx tx L M whenever 0 Thus, by the deﬁnition of a limit,
lim f x xla tx L M x a 2 5E02(pp 098109) 1/19/06 4:19 PM Page 99 SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ❙❙❙❙ 99 I nfinite Limits
Inﬁnite limits can also be deﬁned in a precise way. The following is a precise version of
Deﬁnition 4 in Section 2.2.
6 Definition Let f be a function deﬁned on some open interval that contains the
number a, except possibly at a itself. Then lim f x xla y means that for every positive number M there is a positive number
y=M M 0 x a a∂
FIGURE 10 a+∂ fx M 0 whenever x such that a This says that the values of f x can be made arbitrarily large (larger than any given
number M ) by taking x close enough to a (within a distance , where depends on M , but
with x a). A geometric illustration is shown in Figure 10.
0 such that if we restrict
Given any horizontal line y M , we can ﬁnd a number
x to lie in the interval a
,a
but x a, then the curve y f x lies above the line
y M . You can see that if a larger M is chosen, then a smaller may be required.
EXAMPLE 5 Use Deﬁnition 6 to prove that lim xl0 SOLUTION
1. Guessing a value for . Given M 1
x2 . 0, we want to ﬁnd 1
x2 M whenever 0 x that is, x2 1
M whenever 0 x or x 1
sM whenever 0 0 such that x This suggests that we should take
2. Showing that this works. If M 1 sM.
0 is given, let 0 1 sM . If 0 then Thus 1
x2 M ? x2 2 ? x 1
x2 1 whenever Therefore, by Deﬁnition 6,
lim xl0 1
x2 M 2 0 x 0 x 0 , 5E02(pp 098109) ❙❙❙❙ 100 1/19/06 4:19 PM Page 100 CHAPTER 2 LIMITS AND RATES OF CHANGE y a∂ Similarly, the following is a precise version of Deﬁnition 5 in Section 2.2. It is illustrated by Figure 11. a+∂
a 0 7 Definition Let f be a function deﬁned on some open interval that contains the
number a, except possibly at a itself. Then x lim f x y=N N xla means that for every negative number N there is a positive number such that FIGURE 11 fx  2.4 N whenever 0 x a Exercises 1. How close to 2 do we have to take x so that 5 x 3 is within a y 1 is within a
distance of (a) 0.01, (b) 0.001, and (c) 0.0001 from 29? 2.4
2
1.6 y=œx
„ distance of (a) 0.1 and (b) 0.01 from 13?
2. How close to 5 do we have to take x so that 6 x
3. Use the given graph of f x 1 x to ﬁnd a number such that
0 1
x 0.5 0.2 x whenever ? x ? 4 2
x 2 to ﬁnd a number 6. Use the given graph of f x such that y x2 1
y= x 1 1
2 1 whenever x 1 y 0.7
0.5 y=≈ 1.5 0.3
1
0 10
7 2 4. Use the given graph of f to ﬁnd a number fx 3 0.6 x 10
3 whenever 0.5 such that
x 0 0 5 ? s4x 3.6
3
2.4 1 0.5 3 whenever x 2 ; 8. Use a graph to ﬁnd a number such that  sin x
0 4  0.1 5. Use the given graph of f x 0.4 sx to ﬁnd a number whenever x 4 whenever x 6 ; 9. For the limit x 5 5.7 1
2 lim 4 2 x ? ; 7. Use a graph to ﬁnd a number such that y sx 1 such that xl1 x 3x 3 2 illustrate Deﬁnition 2 by ﬁnding values of
1 and
0.1. that correspond to 5E02(pp 098109) 1/19/06 4:20 PM Page 101 S ECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ; 10. For the limit 21. lim 4x
lim
x l 2 3x xl 5 1
4 4.5 23. lim x
25. lim x 2 that correspond to x 1x 100 2 1 0 whenever x 4x
1 x l2 xl 2 ; 12. For the limit
lim cot 2x ■ ■ 0 xl0 29. lim x 2 1 4
28. lim s9 x xl9 ■ 5 30. lim x 2 1 ■ 32. lim x 3
■ ■ ■ 4 8 8 x l2 ■ 0 x x l3 3 7 3 26. lim x 3 0 12 101 c xla 31. lim x 2 x x
x 24. lim c 27. lim x
xl0 x2 x l3 0 xl0 ; 11. Use a graph to ﬁnd a number such that 22. lim 7 a xla illustrate Deﬁnition 2 by ﬁnding values of
0.5 and
0.1. 2 3x
5 4 ❙❙❙❙ ■ ■ ■ ■ xl0 illustrate Deﬁnition 6 by ﬁnding values of
(a) M 100 and (b) M 1000. 33. Verify that another possible choice of that correspond to lim x l3 x 2 34. Verify, by a geometric argument, that the largest possible 13. A machinist is required to manufacture a circular metal disk with area 1000 cm2.
(a) What radius produces such a disk?
(b) If the machinist is allowed an error tolerance of 5 cm2 in
the area of the disk, how close to the ideal radius in part (a)
must the machinist control the radius?
(c) In terms of the , deﬁnition of lim x l a f x
L, what
is x ? What is f x ? What is a? What is L ? What value of
is given? What is the corresponding value of ? ; 14. A crystal growth furnace is used in research to determine how
best to manufacture crystals used in electronic components for
the space shuttle. For proper growth of the crystal, the temperature must be controlled accurately by adjusting the input power.
Suppose the relationship is given by choice of
CAS for showing that lim x l3 x 2 35. (a) For the limit lim x l 1 x 0.1w Tw 2.155w 36. Prove that lim
x l2 1
x Prove the statement using the ,
illustrate with a diagram like Figure 9. sa if a xla  15. lim 2 x 3 17. lim 1 4x xl1 xl 3 ■ ■ 19–32 ■  19. lim x l3 x
5 16. lim 5 xl 2 ■ ■ ■ ■ x
sx  a
.
sa tion 2.2, prove, using Deﬁnition 2, that lim t l 0 H t does not
exist. [Hint : Use an indirect proof as follows. Suppose that the
1
limit is L. Take
2 in the deﬁnition of a limit and try to
arrive at a contradiction.]
39. If the function f is deﬁned by 0
1 fx if x is rational
if x is irrational prove that lim x l 0 f x does not exist.
40. By comparing Deﬁnitions 2, 3, and 4, prove Theorem 1 in xl6 2 3x ■ 20. lim 3) x xl4 Prove the statement using the ,
3
5 ( 18. lim 7 13 sa 0. Section 2.3.
1
2 41. How close to 5 ■ ■ ■ deﬁnition of limit.
x
4 3 9
2 3. 38. If H is the Heaviside function deﬁned in Example 6 in Sec deﬁnition of limit and  s9 1
.
2 37. Prove that lim s x 20 where T is the temperature in degrees Celsius and w is the
power input in watts.
(a) How much power is needed to maintain the temperature
at 200 C ?
(b) If the temperature is allowed to vary from 200 C by up to
1 C , what range of wattage is allowed for the input
power?
(c) In terms of the , deﬁnition of lim x l a f x
L, what
is x ? What is f x ? What is a? What is L ? What value of
is given? What is the corresponding value of ?
15–18 9 is 3 x1
3, use a graph to ﬁnd a
value of that corresponds to
0.4.
(b) By using a computer algebra system to solve the cubic
equation x 3 x 1 3
, ﬁnd the largest possible
value of that works for any given
0.
(c) Put
0.4 in your answer to part (b) and compare with
your answer to part (a). Hint: Use s x
2 for showing that
min 2, 8 . 9 in Example 4 is 3 do we have to take x so that
1
10,000
x 34 ■ 42. Prove, using Deﬁnition 6, that lim xl 3 43. Prove that lim xl 1 5
x 1 3 . 1
x 34 . 5E02(pp 098109) 102 ❙❙❙❙ 1/19/06 4:20 PM Page 102 CHAPTER 2 LIMITS AND RATES OF CHANGE 44. Suppose that lim x l a f x and lim x l a t x
a real number. Prove each statement.
(a) lim f x tx xla  2.5 c, where c is (b) lim f x t x if c xla 0 if c (c) lim f x t x
xla 0 Continuity
We noticed in Section 2.3 that the limit of a function as x approaches a can often be found
simply by calculating the value of the function at a. Functions with this property are called
continuous at a. We will see that the mathematical deﬁnition of continuity corresponds
closely with the meaning of the word continuity in everyday language. (A continuous
process is one that takes place gradually, without interruption or abrupt change.) Explore continuous functions interactively.
Resources / Module 2
/ Continuity
/ Start of Continuity 1 Definition A function f is continuous at a number a if lim f x
x la  As illustrated in Figure 1, if f is continuous,
then the points x, f x on the graph of f
approach the point a, f a on the graph. So
there is no gap in the curve. fa Notice that Deﬁnition l implicitly requires three things if f is continuous at a:
1. f a is deﬁned (that is, a is in the domain of f )
2. lim f x exists
x la y ƒ
approaches
f(a). 3. lim f x y=ƒ x la f(a) 0 x a As x approaches a,
FIGURE 1 y fa The deﬁnition says that f is continuous at a if f x approaches f a as x approaches a.
Thus, a continuous function f has the property that a small change in x produces only a
small change in f x . In fact, the change in f x can be kept as small as we please by keeping the change in x sufﬁciently small.
If f is deﬁned near a (in other words, f is deﬁned on an open interval containing a,
except perhaps at a), we say that f is discontinuous at a, or f has a discontinuity at a, if
f is not continuous at a.
Physical phenomena are usually continuous. For instance, the displacement or velocity
of a vehicle varies continuously with time, as does a person’s height. But discontinuities
do occur in such situations as electric currents. [See Example 6 in Section 2.2, where the
Heaviside function is discontinuous at 0 because lim t l 0 H t does not exist.]
Geometrically, you can think of a function that is continuous at every number in an
interval as a function whose graph has no break in it. The graph can be drawn without
removing your pen from the paper.
EXAMPLE 1 Figure 2 shows the graph of a function f. At which numbers is f discontinu ous? Why? 0 1 FIGURE 2 2 3 4 5 x SOLUTION It looks as if there is a discontinuity when a
1 because the graph has a break
there. The ofﬁcial reason that f is discontinuous at 1 is that f 1 is not deﬁned.
The graph also has a break when a 3, but the reason for the discontinuity is different. Here, f 3 is deﬁned, but lim x l 3 f x does not exist (because the left and right limits
are different). So f is discontinuous at 3.
What about a 5? Here, f 5 is deﬁned and lim x l 5 f x exists (because the left and
right limits are the same). But
lim f x
f5
xl5 So f is discontinuous at 5. 5E02(pp 098109) 1/19/06 4:20 PM Page 103 S ECTION 2.5 CONTINUITY ❙❙❙❙ 103 Now let’s see how to detect discontinuities when a function is deﬁned by a formula.
EXAMPLE 2 Where are each of the following functions discontinuous? x2 (a) f x Resources / Module 2
/ Continuity
/ Problems and Tests x 2 x 2 x2 x (c) f x 2
2 1 if x 2 if x x 1
x2
1 (b) f x if x 0 if x 0 2 (d) f x x SOLUTION (a) Notice that f 2 is not deﬁned, so f is discontinuous at 2. Later we’ll see why f is
continuous at all other numbers.
(b) Here f 0
1 is deﬁned but
lim f x lim xl0 xl0 1
x2 does not exist. (See Example 8 in Section 2.2.) So f is discontinuous at 0.
(c) Here f 2
1 is deﬁned and
lim f x
x l2 lim
x l2 x2 x
x 2 lim 2 x 2x
x2 x l2 1 lim x
x l2 1 3 exists. But
lim f x f2 x l2 so f is not continuous at 2.
(d) The greatest integer function f x
x has discontinuities at all of the integers
because lim x l n x does not exist if n is an integer. (See Example 10 and Exercise 49 in
Section 2.3.)
Figure 3 shows the graphs of the functions in Example 2. In each case the graph can’t
be drawn without lifting the pen from the paper because a hole or break or jump occurs in
the graph. The kind of discontinuity illustrated in parts (a) and (c) is called removable
because we could remove the discontinuity by redeﬁning f at just the single number 2.
[The function t x
x 1 is continuous.] The discontinuity in part (b) is called an inﬁnite discontinuity. The discontinuities in part (d) are called jump discontinuities because
the function “jumps” from one value to another.
y y y y 1 1 1 1 0 (a) ƒ=
FIGURE 3 1 2 ≈x2
x2 0 x (b) ƒ= Graphs of the functions in Example 2 0 x 1/≈ if x≠0
1
if x=0 (c) ƒ= 1 2 x ≈x2
if x≠2
x2
1
if x=2 0 1 2 (d) ƒ=[ x ] 3 x 5E02(pp 098109) 104 ❙❙❙❙ 1/19/06 4:20 PM Page 104 CHAPTER 2 LIMITS AND RATES OF CHANGE 2 Definition A function f is continuous from the right at a number a if lim f x fa xla and f is continuous from the left at a if
lim f x fa xla EXAMPLE 3 At each integer n, the function f x x [see Figure 3(d)] is continuous
from the right but discontinuous from the left because
lim f x lim x x ln but n x ln lim f x lim x x ln fn n x ln 1 fn 3 Definition A function f is continuous on an interval if it is continuous at
every number in the interval. (If f is deﬁned only on one side of an endpoint of the
interval, we understand continuous at the endpoint to mean continuous from the
right or continuous from the left.) EXAMPLE 4 Show that the function f x interval
S OLUTION If 1 x 2 is continuous on the s1 1, 1 .
1 a 1, then using the Limit Laws, we have
lim (1 lim f x xla x2) s1 xla lim s1 x2 1 s xlim
la 1 x2 1 s1 a2 1 xla (by Laws 2 and 7)
(by 11)
(by 2, 7, and 9) fa
y Thus, by Deﬁnition l, f is continuous at a if
ƒ=1 œ„„„„„
1 ≈ 1 0 FIGURE 4 lim f x xl 1 1 x 1 f 1 1 a and 1. Similar calculations show that
lim f x x l1 1 f1 so f is continuous from the right at 1 and continuous from the left at 1. Therefore,
according to Deﬁnition 3, f is continuous on 1, 1 .
The graph of f is sketched in Figure 4. It is the lower half of the circle
x2 y 1 2 1 Instead of always using Deﬁnitions 1, 2, and 3 to verify the continuity of a function as
we did in Example 4, it is often convenient to use the next theorem, which shows how to
build up complicated continuous functions from simple ones. 5E02(pp 098109) 1/19/06 4:20 PM Page 105 SECTION 2.5 CONTINUITY ❙❙❙❙ 105 4 Theorem If f and t are continuous at a and c is a constant, then the following
functions are also continuous at a :
1. f
t
2. f
t
3. cf
f
if t a
0
4. f t
5.
t Proof Each of the ﬁve parts of this theorem follows from the corresponding Limit Law in Section 2.3. For instance, we give the proof of part 1. Since f and t are continuous at
a, we have
lim f x fa xla and lim t x xla ta Therefore
lim f xla tx lim f x tx xla lim f x lim t x xla fa ta f
This shows that f (by Law 1) xla ta t is continuous at a. It follows from Theorem 4 and Deﬁnition 3 that if f and t are continuous on an interval, then so are the functions f t, f t, c f, f t, and (if t is never 0) f t. The following
theorem was stated in Section 2.3 as the Direct Substitution Property.
5 Theorem (a) Any polynomial is continuous everywhere; that is, it is continuous on
,.
(b) Any rational function is continuous wherever it is deﬁned; that is, it is continuous on its domain.
Proof (a) A polynomial is a function of the form
Px cn x n cn 1 x n 1 c1 x c0 where c0 , c1, . . . , cn are constants. We know that
lim c0 xla and lim x m xla am c0
m (by Law 7) 1, 2, . . . , n (by 9) This equation is precisely the statement that the function f x
x m is a continuous
m
function. Thus, by part 3 of Theorem 4, the function t x
c x is continuous. Since P
is a sum of functions of this form and a constant function, it follows from part 1 of
Theorem 4 that P is continuous. 5E02(pp 098109) 106 ❙❙❙❙ 1/19/06 4:20 PM Page 106 CHAPTER 2 LIMITS AND RATES OF CHANGE (b) A rational function is a function of the form
Px
Qx fx where P and Q are polynomials. The domain of f is D
x
Qx
0 . We know
from part (a) that P and Q are continuous everywhere. Thus, by part 5 of Theorem 4,
f is continuous at every number in D.
As an illustration of Theorem 5, observe that the volume of a sphere varies continuously
4
3
with its radius because the formula V r
3 r shows that V is a polynomial function
of r. Likewise, if a ball is thrown vertically into the air with a velocity of 50 ft s, then the
height of the ball in feet after t seconds is given by the formula h 50t 16t 2. Again this
is a polynomial function, so the height is a continuous function of the elapsed time.
Knowledge of which functions are continuous enables us to evaluate some limits very
quickly, as the following example shows. Compare it with Example 2(b) in Section 2.3.
EXAMPLE 5 Find lim x3 xl 2 2x2 1
.
5 3x SOLUTION The function x3 fx 2x 2 1
5 3x is rational, so by Theorem 5 it is continuous on its domain, which is {x x
Therefore
lim xl 2 x3 2x2 1
5 3x lim f x f xl 2 2 3 5 2 22
32 5
3 }. 2
1 1
11 y P(Ł ¨, Ã ¨)
1
¨
0 (1, 0) x FIGURE 5  Another way to establish the limits in (6) is
to use the Squeeze Theorem with the inequality
sin
0), which is proved in Sec(for
tion 3.5. It turns out that most of the familiar functions are continuous at every number in their
domains. For instance, Limit Law 10 (page 84) implies that root functions are continuous. [Example 3 in Section 2.4 shows that f x
sx is continuous from the right at 0.]
From the appearance of the graphs of the sine and cosine functions (Figure 18 in
Section 1.2), we would certainly guess that they are continuous. We know from the deﬁnitions of sin and cos that the coordinates of the point P in Figure 5 are cos , sin . As
l 0, we see that P approaches the point 1, 0 and so cos l 1 and sin l 0. Thus
6 lim cos
l0 1 lim sin
l0 0 Since cos 0 1 and sin 0 0, the equations in (6) assert that the cosine and sine functions are continuous at 0. The addition formulas for cosine and sine can then be used to
deduce that these functions are continuous everywhere (see Exercises 54 and 55).
It follows from part 5 of Theorem 4 that
tan x sin x
cos x 5E02(pp 098109) 1/19/06 4:21 PM Page 107 S ECTION 2.5 CONTINUITY y ❙❙❙❙ 107 is continuous except where cos x 0. This happens when x is an odd integer multiple of
2, so y tan x has inﬁnite discontinuities when x
2, 3 2, 5 2, and so on
(see Figure 6). 1
3π _π _2 _ π
2 0 π
2 π 3π
2 x 7 Theorem The following types of functions are continuous at every number in
their domains: polynomials
root functions FIGURE 6 rational functions
trigonometric functions y=tan x
EXAMPLE 6 On what intervals is each function continuous? (a) f x x 100 2 x 37 (c) h x sx x
x 75
1
1 x
x2 x2 (b) t x 2 x 17
x2 1 1
1 SOLUTION (a) f is a polynomial, so it is continuous on
,
by Theorem 5(a).
(b) t is a rational function, so by Theorem 5(b), it is continuous on its domain, which is
D
x x2 1 0
xx
1 . Thus, t is continuous on the intervals
, 1,
1, 1 , and 1, .
(c) We can write h x
Fx
Gx
H x , where
Fx sx Gx x
x 1
1 x
x2 Hx 1
1 F is continuous on 0,
by Theorem 7. G is a rational function, so it is continuous
everywhere except when x 1 0, that is, x 1. H is also a rational function, but its
denominator is never 0, so H is continuous everywhere. Thus, by parts 1 and 2 of Theorem 4, h is continuous on the intervals 0, 1 and 1, .
Another way of combining continuous functions f and t to get a new continuous function is to form the composite function f t. This fact is a consequence of the following
theorem.
 This theorem says that a limit symbol can be
moved through a function symbol if the function
is continuous and the limit exists. In other words,
the order of these two symbols can be reversed. 8 Theorem If f is continuous at b and lim t x
x la
In other words, lim f t x xla ( b, then lim f t x f lim t x
xla x la f b. ) Intuitively, Theorem 8 is reasonable because if x is close to a, then t x is close to b,
and since f is continuous at b, if t x is close to b, then f t x is close to f b . A proof
of Theorem 8 is given in Appendix F.
n
Let’s now apply Theorem 8 in the special case where f x
sx , with n being a positive integer. Then
f tx n
st x 5E02(pp 098109) 108 ❙❙❙❙ 1/19/06 4:21 PM Page 108 CHAPTER 2 LIMITS AND RATES OF CHANGE ( f lim t x and xla ) n s xlim t x
la If we put these expressions into Theorem 8, we get
n
lim st x n s xlim t x
la xla and so Limit Law 11 has now been proved. (We assume that the roots exist.) 9 Theorem If t is continuous at a and f is continuous at t a , then the composite
f t x is continuous at a.
function f t given by f t x This theorem is often expressed informally by saying “a continuous function of a continuous function is a continuous function.”
Proof Since t is continuous at a, we have lim t x ta xla Since f is continuous at b t a , we can apply Theorem 8 to obtain
lim f t x f ta xla which is precisely the statement that the function h x
is, f t is continuous at a. f t x is continuous at a; that EXAMPLE 7 Where are the following functions continuous? sin x 2 (a) h x (b) F x 1
sx 2 7 4 SOLUTION (a) We have h x f t x , where
tx x2 and fx sin x Now t is continuous on since it is a polynomial, and f is also continuous everywhere.
Thus, h f t is continuous on by Theorem 9.
(b) Notice that F can be broken up as the composition of four continuous functions:
F
where fx fthk 1
x tx x Fx or
4 hx f thkx
sx kx x2 7 We know that each of these functions is continuous on its domain (by Theorems 5 and 7),
so by Theorem 9, F is continuous on its domain, which is {x sx 2 7 4} xx 3 , 3 3, 3 3, An important property of continuous functions is expressed by the following theorem,
whose proof is found in more advanced books on calculus. 5E02(pp 098109) 1/19/06 4:21 PM Page 109 SECTION 2.5 CONTINUITY ❙❙❙❙ 109 10 The Intermediate Value Theorem Suppose that f is continuous on the closed interval a, b and let N be any number between f a and f b , where f a
f b.
Then there exists a number c in a, b such that f c
N. The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values f a and f b . It is illustrated by Figure 7. Note
that the value N can be taken on once [as in part (a)] or more than once [as in part (b)].
y y f(b) f(b) y=ƒ N
N y=ƒ f(a)
0 a f(a) y=ƒ
y=N N
f(b)
0 a b x 0 x cb FIGURE 7
y f(a) a c¡ c™ (a) c£ b x (b) If we think of a continuous function as a function whose graph has no hole or break,
then it is easy to believe that the Intermediate Value Theorem is true. In geometric terms it
says that if any horizontal line y N is given between y f a and y f b as in Figure 8, then the graph of f can’t jump over the line. It must intersect y N somewhere.
It is important that the function f in Theorem 10 be continuous. The Intermediate Value
Theorem is not true in general for discontinuous functions (see Exercise 42).
One use of the Intermediate Value Theorem is in locating roots of equations as in the
following example. FIGURE 8 EXAMPLE 8 Show that there is a root of the equation 4x 3 6x 2 3x 2 0 between 1 and 2.
4 x 3 6 x 2 3x 2. We are looking for a solution of the given
equation, that is, a number c between 1 and 2 such that f c
0. Therefore, we take
a 1, b 2, and N 0 in Theorem 10. We have
SOLUTION Let f x f1
f2 and 4 6 32 24 3 2
6 1
2 12 0
0 Thus, f 1
0 f 2 ; that is, N 0 is a number between f 1 and f 2 . Now f is
continuous since it is a polynomial, so the Intermediate Value Theorem says there
is a number c between 1 and 2 such that f c
0. In other words, the equation
4 x 3 6 x 2 3x 2 0 has at least one root c in the interval 1, 2 .
In fact, we can locate a root more precisely by using the Intermediate Value Theorem
again. Since
f 1.2 0.128 0 and f 1.3 0.548 0 5E02(pp 110121) 110 ❙❙❙❙ 1/17/06 12:29 PM Page 110 CHAPTER 2 LIMITS AND RATES OF CHANGE a root must lie between 1.2 and 1.3. A calculator gives, by trial and error,
f 1.22 0.007008 0 and f 1.23 0.056068 0 so a root lies in the interval 1.22, 1.23 .
We can use a graphing calculator or computer to illustrate the use of the Intermediate
Value Theorem in Example 8. Figure 9 shows the graph of f in the viewing rectangle
1, 3 by 3, 3 and you can see that the graph crosses the xaxis between 1 and 2. Figure 10 shows the result of zooming in to the viewing rectangle 1.2, 1.3 by 0.2, 0.2 .
3 0.2 3 _1 1.3 1.2 _3 _0.2 FIGURE 9 FIGURE 10 In fact, the Intermediate Value Theorem plays a role in the very way these graphing
devices work. A computer calculates a ﬁnite number of points on the graph and turns on
the pixels that contain these calculated points. It assumes that the function is continuous
and takes on all the intermediate values between two consecutive points. The computer
therefore connects the pixels by turning on the intermediate pixels.  2.5 Exercises 1. Write an equation that expresses the fact that a function f y is continuous at the number 4.
2. If f is continuous on , , what can you say about its graph?
3. (a) From the graph of f , state the numbers at which f is discontinuous and explain why.
(b) For each of the numbers stated in part (a), determine
whether f is continuous from the right, or from the left,
or neither.
y _4 _2 2 4 6 8 x 5. Sketch the graph of a function that is continuous everywhere except at x 3 and is continuous from the left at 3. 6. Sketch the graph of a function that has a jump discontinuity at x 2 and a removable discontinuity at x
ous elsewhere.
_4 _2 0 2 4 6 4. From the graph of t, state the intervals on which t is continuous. x 4, but is continu 7. A parking lot charges $3 for the ﬁrst hour (or part of an hour) and $2 for each succeeding hour (or part), up to a daily maximum of $10.
(a) Sketch a graph of the cost of parking at this lot as a function of the time parked there.
(b) Discuss the discontinuities of this function and their
signiﬁcance to someone who parks in the lot. 5E02(pp 110121) 1/17/06 12:29 PM Page 111 ❙❙❙❙ S ECTION 2.5 CONTINUITY 8. Explain why each function is continuous or discontinuous. (a) The temperature at a speciﬁc location as a function of time
(b) The temperature at a speciﬁc time as a function of the distance due west from New York City
(c) The altitude above sea level as a function of the distance
due west from New York City
(d) The cost of a taxi ride as a function of the distance traveled
(e) The current in the circuit for the lights in a room as a function of time
9. If f and t are continuous functions with f 3 lim x l 3 2 f x 5 and 10–12  Use the deﬁnition of continuity and the properties of limits to show that the function is continuous at the given number.
10. f x x2 11. f x x 12. t x s7 x, a 25. h x cos 1 27. F x 1 2 x 3 4, x1
,a
2x 2 1 ■ ; 29–30 ■ 13–14 ■ ■ ■ ■ ■ ■ Use the deﬁnition of continuity and the properties of limits to show that the function is continuous on the given interval.
2x
x 3
,
2 14. t x 2 s3 x, ■ ■ ■ ■ ■ 31–32 ■ ■ ■ ■ ■ ■ ■ 15–20  Explain why the function is discontinuous at the given
number a. Sketch the graph of the function. 1
1 x
2 a 2 1
16. f x if x 1 1 x2
1x 18. f x x2
x2
1 19. f x if x
if x
if x ■ ■ ■ 1 x2 x a 1 ■ ■ ■ tan sx
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 5 sx
s5 x
sin x ■ ■ ■ ■ ■ ■ 33–34  ■ Show that f is continuous on
2 33. f x x if x
sx if x
sin x if x
cos x if x , . 1
1 ■ ■ ■ 12
3 if x 3
3 x 2 if x
x if x ■ ■ 6 ■ ■ ■ 1
2
x x2
x
2 2 if x
if 0
if x 0
x
2 2 1
x
3 3 x 2 if x
2x 2
if 0
2 x if x 0
x
1 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 3
38. The gravitational force exerted by Earth on a unit mass at a dis tance r from the center of the planet is 1
1 ■ ■ x1
if x
1x
if 1
sx 3 if x ■ a ■ 4
4 37. f x 1 if x x
5x ■ 36. f x a
■ ■ ■ 1
■ ■ ■ 21–28  Explain, using Theorems 4, 5, 7, and 9, why the function
is continuous at every number in its domain. State the domain.
21. F x ■ Use continuity to evaluate the limit. 35. f x 1 a 1 5
1
4 ■ 35–37  Find the numbers at which f is discontinuous. At which
of these numbers is f continuous from the right, from the left, or
neither? Sketch the graph of f . 1 a 1
1 if x
x  1 x
1 x2 20. f x 1 if x 17. f x ■ 30. y 34. f x ,3
■ x ■ 2, ■ 15. f x ■ ■ ■  13. f x sin cos sin x ■ ■ 1
sin x 1 xl ■ ■ Locate the discontinuities of the function and illustrate by  x l4 ■ tan 2 x 28. F x ■ 32. lim sin x
■ x2 graphing. ■
■ 1 sin x
x1 26. h x s2 x ■ 31. lim 4 24. h x sx sin x ■ 4 a x2 29. y 4, ﬁnd t 3 . tx 23. R x 111 22. G x 3
sx 1 x3 Fr GMr
R3
GM
r2 if r R if r R where M is the mass of Earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r ? 5E02(pp 110121) 112 ❙❙❙❙ 1/17/06 12:30 PM Page 112 CHAPTER 2 LIMITS AND RATES OF CHANGE ; 51–52 39. For what value of the constant c is the function f continuous , on  (a) Prove that the equation has at least one real root.
(b) Use your graphing device to ﬁnd the root correct to three decimal places.
1
51. x 5
x2 4 0
52. s x
5
x3 ?
cx
cx 2 fx 1
1 if x
if x 3
3 40. Find the constant c that makes t continuous on , . ■ 2 x
cx tx 2 c
20 if x
if x 4
4 ■ ■ ■ ■ lim f a lim sin a 47. cos x
■ ■ x, 0, ■ 1 ■ ■ ■ 2 x,
■ ■ ■ ■ 2
■  2.6 50. x 5 x
■ ■ ■ ■ ■ x2
■ 2x
■ 3 ■ 0
■ if x is rational
if x is irrational 58. For what values of x is t continuous? 0
x if x is rational
if x is irrational 60. (a) Show that the absolute value function F x 0, 1.4
■ 0
1 59. Is there a number that is exactly 1 more than its cube? 49–50  (a) Prove that the equation has at least one real root.
(b) Use your calculator to ﬁnd an interval of length 0.01 that contains a root.
49. sin x fx tx 0, 1 x, 48. tan x
■ sin a 57. For what values of x is f continuous? 2. (This proves the existence 3
46. sx 0, 1 ■ fa (b) Prove Theorem 4, part 5. x, show that there is a number c such 1, 2 ■ 56. (a) Prove Theorem 4, part 3. 45–48  Use the Intermediate Value Theorem to show that there is
a root of the given equation in the speciﬁed interval. 3 ■ 55. Prove that cosine is a continuous function. positive number c such that c 2
of the number s2.) x ■ Use (6) to show that this is true. 44. Use the Intermediate Value Theorem to prove that there is a 45. x 4 h hl0 0.25 and that f 0
1 and f 1
3. Let N 2. Sketch two
possible graphs of f , one showing that f might not satisfy the
conclusion of the Intermediate Value Theorem and one showing that f might still satisfy the conclusion of the Intermediate
Value Theorem (even though it doesn’t satisfy the hypothesis).
x3 x2
10. ■ lim x l a sin x sin a for every real number a. By Exercise 53
an equivalent statement is that 42. Suppose that a function f is continuous on [0, 1] except at that f c ■ 54. To prove that sine is continuous, we need to show that x 3 64
,a
4
x4
3 sx
,a 9
9x 43. If f x h hl0 nuity at a ? If the discontinuity is removable, ﬁnd a function t
that agrees with f for x a and is continuous on .
x 2 2x 8
(a) f x
,a
2
x2
x7
(b) f x
,a 7
x7 (d) f x ■ 53. Prove that f is continuous at a if and only if 41. Which of the following functions f has a removable disconti (c) f x ■ ■ x is continuous everywhere.
(b) Prove that if f is a continuous function on an interval, then
so is f .
(c) Is the converse of the statement in part (b) also true? In
other words, if f is continuous, does it follow that f is
continuous? If so, prove it. If not, ﬁnd a counterexample. 61. A Tibetan monk leaves the monastery at 7:00 A.M. and takes his usual path to the top of the mountain, arriving at 7:00 P.M.
The following morning, he starts at 7:00 A.M. at the top and
takes the same path back, arriving at the monastery at 7:00 P.M.
Use the Intermediate Value Theorem to show that there is a
point on the path that the monk will cross at exactly the same
time of day on both days. Tangents, Velocities, and Other Rates of Change
In Section 2.1 we guessed the values of slopes of tangent lines and velocities on the basis
of numerical evidence. Now that we have deﬁned limits and have learned techniques for
computing them, we return to the tangent and velocity problems with the ability to calculate slopes of tangents, velocities, and other rates of change. 5E02(pp 110121) 1/17/06 12:30 PM Page 113 ❙❙❙❙ S ECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE 113 Tangents
If a curve C has equation y f x and we want to ﬁnd the tangent line to C at the point
P a, f a , then we consider a nearby point Q x, f x , where x a, and compute the slope
of the secant line PQ :
fx
x mPQ fa
a Then we let Q approach P along the curve C by letting x approach a. If mPQ approaches a
number m, then we deﬁne the tangent t to be the line through P with slope m. (This
amounts to saying that the tangent line is the limiting position of the secant line PQ as Q
approaches P. See Figure 1.)
y y t
Q Q{ x, ƒ } Q ƒf(a) Q P P { a, f(a)}
xa 0 a x x x 0 FIGURE 1 1 Definition The tangent line to the curve y
line through P with slope m lim xla f x at the point P a, f a is the fx
x fa
a provided that this limit exists.
In our ﬁrst example we conﬁrm the guess we made in Example 1 in Section 2.1.
x 2 at the EXAMPLE 1 Find an equation of the tangent line to the parabola y point P 1, 1 .
SOLUTION Here we have a m lim
x l1 fx
x lim x x l1 lim x
x l1  Pointslope form for a line through the
point x1 , y1 with slope m :
y y1 mx x 2, so the slope is 1 and f x lim
x l1 f1
1
1x
x1
1 x2
x 1 2 1
1 1
1 Using the pointslope form of the equation of a line, we ﬁnd that an equation of the
tangent line at 1, 1 is x1 y 1 2x 1 or y 2x 1 5E02(pp 110121) 114 ❙❙❙❙ 1/17/06 12:30 PM Page 114 CHAPTER 2 LIMITS AND RATES OF CHANGE We sometimes refer to the slope of the tangent line to a curve at a point as the slope of
the curve at the point. The idea is that if we zoom in far enough toward the point, the curve
looks almost like a straight line. Figure 2 illustrates this procedure for the curve y x 2 in
Example 1. The more we zoom in, the more the parabola looks like a line. In other words,
the curve becomes almost indistinguishable from its tangent line.
2 1.5 1.1 (1, 1) 0 (1, 1) 2 (1, 1) 1.5 0.5 1.1 0.9 FIGURE 2 Zooming in toward the point (1, 1) on the parabola y=≈ There is another expression for the slope of a tangent line that is sometimes easier to
use. Let
h a x Then x
a h so the slope of the secant line PQ is
fa mPQ h
h fa (See Figure 3 where the case h 0 is illustrated and Q is to the right of P. If it happened
that h 0, however, Q would be to the left of P.)
y t
Q{ a+h, f(a+h)}
f(a+h)f(a) P { a, f(a)}
h 0 a+h a x FIGURE 3 Notice that as x approaches a, h approaches 0 (because h
sion for the slope of the tangent line in Deﬁnition 1 becomes 2 m lim hl0 fa h
h x a) and so the expres fa EXAMPLE 2 Find an equation of the tangent line to the hyperbola y
point 3, 1 . 3 x at the 5E02(pp 110121) 1/17/06 12:30 PM Page 115 S ECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE SOLUTION Let f x f3 lim h
h hl0 3
3 lim lim hl0 f3
3 1 h
h hl0 h
h 1 lim h h 3 lim hl0 h
h3 3 3 hl0 h 1
3 y y= 3
x Therefore, an equation of the tangent at the point 3, 1 is (3, 1)
0 115 3 x. Then the slope of the tangent at 3, 1 is
m x+3y6=0 ❙❙❙❙ y x which simpliﬁes to
FIGURE 4 1
3 1
x 3y x 3 6 0 The hyperbola and its tangent are shown in Figure 4.
EXAMPLE 3 Find the slopes of the tangent lines to the graph of the function f x sx at the points (1, 1), (4, 2), and (9, 3).
SOLUTION Since three slopes are requested, it is efﬁcient to start by ﬁnding the slope at the
general point (a, sa ): m lim lim Rationalize the numerator fa hl0 sa hl0 lim hl0 lim Continuous function of h hl0 h
h
h
h a
h(sa
sa At the point (1, 1), we have a
At (4, 2), we have m 1 (2 s4 ) sa lim h
h hl0 sa h
h
1
h fa sa
sa a
sa ) h
h
lim hl0 sa
sa
h(sa 1
sa sa sa sa h
h sa ) 1
2 sa 1, so the slope of the tangent is m
1
1 (2 s9 ) 1 .
4 ; at (9, 3), m
6 1 (2 s1 ) 1
2 . Velocities
Learn about average and instantaneous velocity
by comparing falling objects.
Resources / Module 3
/ Derivative at a Point
/ The Falling Robot In Section 2.1 we investigated the motion of a ball dropped from the CN Tower and deﬁned
its velocity to be the limiting value of average velocities over shorter and shorter time
periods.
In general, suppose an object moves along a straight line according to an equation
of motion s f t , where s is the displacement (directed distance) of the object from the 5E02(pp 110121) 116 ❙❙❙❙ 1/17/06 12:31 PM Page 116 CHAPTER 2 LIMITS AND RATES OF CHANGE origin at time t. The function f that describes the motion is called the position function
of the object. In the time interval from t a to t a h the change in position is
fa h
f a . (See Figure 5.) The average velocity over this time interval is
displacement
time average velocity fa h
h fa which is the same as the slope of the secant line PQ in Figure 6.
s Q { a+h, f(a+h)}
P { a, f(a)} position at
time t=a position at
time t=a+h h s 0 0 f(a+h)f(a) a f(a) mPQ=
f(a+h) a+h t f(a+h) (a)
f
h average velocity FIGURE 5 FIGURE 6 Now suppose we compute the average velocities over shorter and shorter time intervals
a, a h . In other words, we let h approach 0. As in the example of the falling ball, we
deﬁne the velocity (or instantaneous velocity) v a at time t a to be the limit of these
average velocities:
va 3 fa lim hl0 h
h fa This means that the velocity at time t a is equal to the slope of the tangent line at P
(compare Equations 2 and 3).
Now that we know how to compute limits, let’s reconsider the problem of the falling ball.
EXAMPLE 4 Suppose that a ball is dropped from the upper observation deck of the
CN Tower, 450 m above the ground.
(a) What is the velocity of the ball after 5 seconds?
(b) How fast is the ball traveling when it hits the ground?
 Recall from Section 2.1: The distance
(in meters) fallen after t seconds is 4.9t 2. SOLUTION We ﬁrst use the equation of motion s ft 4.9t 2 to ﬁnd the velocity v a after a seconds:
va lim fa hl0 4.9 a 2 h
h fa hl0 lim 4.9 2a h hl0 (a) The velocity after 5 s is v 5 4.9 a hl0 h2 2ah
h lim lim a2 lim hl0 9.8a 9.8 5 49 m s. h2
h 4.9a 2 4.9 2ah
h h2 5E02(pp 110121) 1/17/06 12:31 PM Page 117 S ECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE ❙❙❙❙ 117 (b) Since the observation deck is 450 m above the ground, the ball will hit the ground at
the time t1 when s t1
450, that is,
4.9t 2
1 450 This gives
t2
1 450
4.9 and 450
4.9 t1 9.6 s The velocity of the ball as it hits the ground is therefore
v t1 9.8t1 9.8 450
4.9 94 m s Other Rates of Change
Suppose y is a quantity that depends on another quantity x. Thus, y is a function of x and
we write y f x . If x changes from x 1 to x 2 , then the change in x (also called the increment of x) is
x x2 x1 and the corresponding change in y is
y f x2 f x1 y
x f x2
x2 f x1
x1 The difference quotient is called the average rate of change of y with respect to x over the interval x 1, x 2 and
can be interpreted as the slope of the secant line PQ in Figure 7.
y Q{ ¤, ‡}
Îy P { ⁄, ﬂ}
Îx
0 FIGURE 7 ⁄ ¤ x average rate of change mPQ
instantaneous rate of change slope of tangent at P By analogy with velocity, we consider the average rate of change over smaller and
smaller intervals by letting x 2 approach x 1 and therefore letting x approach 0. The limit
of these average rates of change is called the (instantaneous) rate of change of y with
respect to x at x x 1 , which is interpreted as the slope of the tangent to the curve y f x
at P x 1, f x 1 : 5E02(pp 110121) 118 ❙❙❙❙ 1/17/06 12:31 PM Page 118 CHAPTER 2 LIMITS AND RATES OF CHANGE instantaneous rate of change 4 xh TC xh TC 0
1
2
3
4
5
6
7
8
9
10
11
12 . 6.5
6.1
5.6
4.9
4.2
4.0
4.0
4.8
6.1
8.3
10.0
12.1
14.3 13
14
15
16
17
18
19
20
21
22
23
24 16.0
17.3
18.2
18.8
17.6
16.0
14.1
11.5
10.2
9.0
7.9
7.0 . xl0 lim x2 l x1 f x2
x2 f x1
x1 E XAMPLE 5 Temperature readings T (in degrees Celsius) were recorded every hour starting at midnight on a day in April in Whiteﬁsh, Montana. The time x is measured in hours
from midnight. The data are given in the table at the left.
(a) Find the average rate of change of temperature with respect to time
(i) from noon to 3 P.M.
(ii) from noon to 2 P.M.
(iii) from noon to 1 P.M.
(b) Estimate the instantaneous rate of change at noon.
SOLUTION (a) (i) From noon to 3 P.M. the temperature changes from 14.3°C to 18.2°C, so
T T 15 T 12 18.2 14.3 3.9 C while the change in time is x 3 h. Therefore, the average rate of change of
temperature with respect to time is
T
x  A NOTE ON UNITS
The units for the average rate of change
T x are the units for T divided by the
units for x, namely, degrees Celsius per hour.
The instantaneous rate of change is the limit of
the average rates of change, so it is measured
in the same units: degrees Celsius per hour. y
x lim 3.9
3 1.3 C h (ii) From noon to 2 P.M. the average rate of change is
T
x T 14
14 T 12
12 17.3 14.3
2 1.5 C h (iii) From noon to 1 P.M. the average rate of change is
T
x T 13
13 T 12
12 16.0 14.3
1 1.7 C h (b) We plot the given data in Figure 8 and use them to sketch a smooth curve that
approximates the graph of the temperature function.
T B 18
16 P 14
12
10 A 8
6
4
2 FIGURE 8 0 1 C 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Then we draw the tangent at the point P where x x 12 and, after measuring the sides of 5E02(pp 110121) 1/17/06 12:31 PM Page 119 SECTION 2.6 TANGENTS, VELOCITIES, AND OTHER RATES OF CHANGE ❙❙❙❙ 119 triangle ABC, we estimate that the slope of the tangent line is
BC
AC  Another method is to average the slopes of
two secant lines. See Example 2 in Section 2.1. 10.3
5.5 1.9 Therefore, the instantaneous rate of change of temperature with respect to time at noon
is about 1.9°C h.
The velocity of a particle is the rate of change of displacement with respect to time.
Physicists are interested in other rates of change as well—for instance, the rate of change
of work with respect to time (which is called power). Chemists who study a chemical reaction are interested in the rate of change in the concentration of a reactant with respect to
time (called the rate of reaction). A steel manufacturer is interested in the rate of change
of the cost of producing x tons of steel per day with respect to x (called the marginal cost).
A biologist is interested in the rate of change of the population of a colony of bacteria with
respect to time. In fact, the computation of rates of change is important in all of the natural sciences, in engineering, and even in the social sciences. Further examples will be
given in Section 3.4.
All these rates of change can be interpreted as slopes of tangents. This gives added signiﬁcance to the solution of the tangent problem. Whenever we solve a problem involving
tangent lines, we are not just solving a problem in geometry. We are also implicitly solving a great variety of problems involving rates of change in science and engineering.  2.6 Exercises 1. A curve has equation y 5. (a) Find the slope of the tangent line to the parabola f x.
(a) Write an expression for the slope of the secant line through
the points P 3, f 3 and Q x, f x .
(b) Write an expression for the slope of the tangent line at P. 2. Suppose an object moves with position function s f t.
(a) Write an expression for the average velocity of the object in
the time interval from t a to t a h.
(b) Write an expression for the instantaneous velocity at
time t a. y x 2 2 x at the point 3, 3
(i) using Deﬁnition 1
(ii) using Equation 2
(b) Find an equation of the tangent line in part (a).
(c) Graph the parabola and the tangent line. As a check on your
work, zoom in toward the point ( 3, 3) until the parabola
and the tangent line are indistinguishable. ; 6. (a) Find the slope of the tangent line to the curve y point 1, 1
(i) using Deﬁnition 1
(ii) using Equation 2
(b) Find an equation of the tangent line in part (a).
(c) Graph the curve and the tangent line in successively smaller
viewing rectangles centered at ( 1, 1) until the curve and
the line appear to coincide. 3. Consider the slope of the given curve at each of the ﬁve points shown. List these ﬁve slopes in decreasing order and explain
your reasoning.
y ;
A E B 7–10  Find an equation of the tangent line to the curve at the
given point. D 7. y ; 4. Graph the curve y sin x in the viewing rectangles 2, 2
by 2, 2 , 1, 1 by 1, 1 , and 0.5, 0.5 by 0.5, 0.5 .
What do you notice about the curve as you zoom in toward the
origin? s2 x 9. y x 1 8. y C
0 x 3 at the x 10. y
■ 1,
1 2x x
■ x 3, 2x ■ x 1, 2 4, 3
2, 1 2,
■ 3, 2 0, 0
■ ■ ■ ■ ■ ■ ■ ■ 5E02(pp 110121) 120 ❙❙❙❙ 1/17/06 12:32 PM Page 120 CHAPTER 2 LIMITS AND RATES OF CHANGE 11. (a) Find the slope of the tangent to the curve y 2 x 3 at
the point where x a.
(b) Find the slopes of the tangent lines at the points whose
xcoordinates are (i) 1, (ii) 0, and (iii) 1. 12. (a) Find the slope of the tangent to the parabola ; y 1 x x 2 at the point where x a.
(b) Find the slopes of the tangent lines at the points whose
xcoordinates are (i) 1, (ii) 1, and (iii) 1.
2
(c) Graph the curve and the three tangents on a common
screen.
x 3 4x 1
at the point where x a.
(b) Find equations of the tangent lines at the points 1, 2
and 2, 1 .
(c) Graph the curve and both tangents on a common screen. 13. (a) Find the slope of the tangent to the curve y ; 14. (a) Find the slope of the tangent to the curve y ; 1 sx at the
point where x a.
(b) Find equations of the tangent lines at the points 1, 1 and
(4, 1 ).
2
(c) Graph the curve and both tangents on a common screen. 15. The graph shows the position function of a car. Use the shape of the graph to explain your answers to the following
questions.
(a) What was the initial velocity of the car?
(b) Was the car going faster at B or at C ?
(c) Was the car slowing down or speeding up at A, B, and C ?
(d) What happened between D and E ?
s D E C
B
A 18. If an arrow is shot upward on the moon with a velocity of 58 m s, its height (in meters) after t seconds is given by
H 58 t 0.83t 2.
(a) Find the velocity of the arrow after one second.
(b) Find the velocity of the arrow when t a.
(c) When will the arrow hit the moon?
(d) With what velocity will the arrow hit the moon?
19. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s 4 t 3 6 t 2,
where t is measured in seconds. Find the velocity of the particle at times t a, t 1, t 2, and t 3.
20. The displacement (in meters) of a particle moving in a straight line is given by s t 2 8 t 18, where t is measured in
seconds.
(a) Find the average velocity over each time interval:
(i) 3, 4
(ii) 3.5, 4
(iii) 4, 5
(iv) 4, 4.5
(b) Find the instantaneous velocity when t 4.
(c) Draw the graph of s as a function of t and draw the secant
lines whose slopes are the average velocities in part (a) and
the tangent line whose slope is the instantaneous velocity in
part (b).
21. A warm can of soda is placed in a cold refrigerator. Sketch the graph of the temperature of the soda as a function of time. Is
the initial rate of change of temperature greater or less than the
rate of change after an hour?
22. A roast turkey is taken from an oven when its temperature has reached 185°F and is placed on a table in a room where the
temperature is 75°F. The graph shows how the temperature of
the turkey decreases and eventually approaches room temperature. (In Section 10.4 we will be able to use Newton’s Law of
Cooling to ﬁnd an equation for T as a function of time.) By
measuring the slope of the tangent, estimate the rate of change
of the temperature after an hour.
T (°F) 0 t 200 P
16. Valerie is driving along a highway. Sketch the graph of the position function of her car if she drives in the following manner: At time t 0, the car is at mile marker 15 and is traveling
at a constant speed of 55 mi h. She travels at this speed for
exactly an hour. Then the car slows gradually over a 2minute
period as Valerie comes to a stop for dinner. Dinner lasts
26 min; then she restarts the car, gradually speeding up to
65 mi h over a 2minute period. She drives at a constant
65 mi h for two hours and then over a 3minute period gradually slows to a complete stop.
17. If a ball is thrown into the air with a velocity of 40 ft s, its height (in feet) after t seconds is given by y
Find the velocity when t 2. 40 t 16 t 2. 100 0 30 60 90 120 150 t (min) 23. (a) Use the data in Example 5 to ﬁnd the average rate of change of temperature with respect to time
(i) from 8 P.M. to 11 P.M.
(ii) from 8 P.M. to 10 P.M.
(iii) from 8 P.M. to 9 P.M.
(b) Estimate the instantaneous rate of change of T with respect
to time at 8 P.M. by measuring the slope of a tangent. 5E02(pp 110121) 1/17/06 12:32 PM Page 121 C HAPTER 2 REVIEW 24. The population P (in thousands) of Belgium from 1992 to 2000 Year Year 1992 1994 1996 1998 2000 P 10,036 10,109 10,152 10,175 10,186 (a) Find the average rate of growth
(i) from 1992 to 1996
(ii) from 1994 to 1996
(iii) from 1996 to 1998
In each case, include the units.
(b) Estimate the instantaneous rate of growth in 1996 by
taking the average of two average rates of change. What
are its units?
(c) Estimate the instantaneous rate of growth in 1996 by measuring the slope of a tangent. 1994 1995 1996 304 572 873 1513 2461 (a) Find the average rate of growth
(i) from 1995 to 1997
(ii) from 1995 to 1996
(iii) from 1994 to 1995
In each case, include the units.
(b) Estimate the instantaneous rate of growth in 1995 by
taking the average of two average rates of change. What
are its units?
(c) Estimate the instantaneous rate of growth in 1995 by measuring the slope of a tangent. ■ sketch.
(a) lim f x
x la (c) lim f x
x la L
L (b) lim f x
x la L (d) lim f x
x la (e) lim f x 1886 2135 3300 100,000 1 t
60 2 0 t 60 Find the rate at which the water is ﬂowing out of the tank (the
instantaneous rate of change of V with respect to t ) as a function of t. What are its units? For times t 0, 10, 20, 30, 40, 50,
and 60 min, ﬁnd the ﬂow rate and the amount of water remaining in the tank. Summarize your ﬁndings in a sentence or two.
At what time is the ﬂow rate the greatest? The least? ■ 4. State the following Limit Laws. (a)
(c)
(e)
(g) Sum Law
Constant Multiple Law
Quotient Law
Root Law (b) Difference Law
(d) Product Law
(f ) Power Law 5. What does the Squeeze Theorem say? xla 2. Describe several ways in which a limit can fail to exist. Illustrate with sketches.
3. What does it mean to say that the line x asymptote of the curve y
various possibilities. 1412 Vt CONCEPT CHECK 1. Explain what each of the following means and illustrate with a 1015 be drained from the bottom of the tank in an hour, then Torricelli’s Law gives the volume V of water remaining in the tank
after t minutes as given in the table. (The numbers of locations as of June 30
are given.) 2 Review 2000 28. If a cylindrical tank holds 100,000 gallons of water, which can 26. The number N of locations of a popular coffeehouse chain is  1999 modity is C x
5000 10 x 0.05x 2.
(a) Find the average rate of change of C with respect to x when
the production level is changed
(i) from x 100 to x 105
(ii) from x 100 to x 101
(b) Find the instantaneous rate of change of C with respect to x
when x 100. (This is called the marginal cost. Its signiﬁcance will be explained in Section 3.4.) 1997 N 1998 27. The cost (in dollars) of producing x units of a certain com Malaysia is shown in the table. (Midyear estimates are given.)
1993 1997 121 (a) Find the average rate of growth
(i) from 1996 to 1998
(ii) from 1997 to 1998
(iii) from 1998 to 1999
In each case, include the units.
(b) Estimate the instantaneous rate of growth in 1998 by
taking the average of two average rates of change. What
are its units?
(c) Estimate the instantaneous rate of growth in 1998 by measuring the slope of a tangent. 25. The number N (in thousands) of cellular phone subscribers in Year 1996 N is shown in the table. (Midyear estimates are given.) ❙❙❙❙ a is a vertical
f x ? Draw curves to illustrate the 6. (a) What does it mean for f to be continuous at a ? (b) What does it mean for f to be continuous on the interval
, ? What can you say about the graph of such a
function?
7. What does the Intermediate Value Theorem say? 5E02(pp 122123) 122 ❙❙❙❙ 1/17/06 12:23 PM Page 122 CHAPTER 2 LIMITS AND RATES OF CHANGE 8. Write an expression for the slope of the tangent line to the curve y 10. If y f x and x changes from x 1 to x 2 , write expressions for
the following.
(a) The average rate of change of y with respect to x over the
interval x 1, x 2 .
(b) The instantaneous rate of change of y with respect to x
at x x 1. f x at the point a, f a . 9. Suppose an object moves along a straight line with position f t at time t. Write an expression for the instantaneous velocity of the object at time t a. How can you interpret this
velocity in terms of the graph of f ? ■ TRUEFALSE QUIZ
7. If p is a polynomial, then lim x l b p x 1. lim
x l4 2. lim
x l1 3. lim
x l1 8 x 4 x2
x2 6x
5x
x x2 x
7
6 lim 4 x lim x 2 6x lim x 5x 6 x l1 x l1
2 lim x
x l1 4 x l4 x 4 2x tx and lim x l 0 t x
0. 9. If the line x 1 is a vertical asymptote of y
not deﬁned at 1. 0 and f 3
0, then there exists a number c
between 1 and 3 such that f c
0. 11. If f is continuous at 5 and f 5 lim x l 2 f 4 x 2
4 2 and lim x l 5 t x
f x t x does not exist. 5. If lim x l 5 f x 0 and lim x l 5 t x
lim x l 5 f x t x does not exist. f x , then f is 10. If f 1 3 4. If lim x l 5 f x limx l 5 , then lim x l 0 f x 8 7 2 x l1 lim x
4 lim x l4 3
2x 2x pb. 8. If lim x l 0 f x Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. 2x ■ 11 2 and f 4 12. If f is continuous on 1, 1 and f 1
then there exists a number r such that r 0, then 4 and f 1
1 and f r 13. Let f be a function such that lim x l 0 f x exists a number such that if 0
fx
6
1. 0, then 3, then 2.
3,
. 6. Then there
, then x 14. If f x 1 for all x and lim x l 0 f x exists, then
lim x l 0 f x
1. 6. If lim x l 6 f x t x exists, then the limit must be f 6 t 6 . ■ EXERCISES ■ (b) State the equations of the vertical asymptotes.
(c) At what numbers is f discontinuous? Explain. 1. The graph of f is given.
y 2. Sketch the graph of an example of a function f that satisﬁes all of the following conditions:
lim f x
2, lim f x
x l0 1 x l0 , lim f x x l2 0  (a) Find each limit, or explain why it does not exist.
(i) lim f x
(ii) lim f x
xl 3 (iii) lim f x (iv) lim f x (v) lim f x (vi) lim f x x l0 lim f x x l2 Find the limit. 3. lim cos x xl 3 1, f0 x 1 3–16 x l2 1, xl0 5. lim xl 3 x l4 x l2 sin x 7. lim h l0 x
x2
h 2 9
2x 13
h 3
1 4. lim
x l3 6. lim
x l1 8. lim
t l2 x2
x 2 9
2x x 2 x2
t2
t3 9
2x 4
8 3
3 5E02(pp 122123) 1/17/06 12:24 PM Page 123 C HAPTER 2 REVIEW 9. lim r r l9 sr
9 4
11. lim
s l16 s x l8 15. lim 12. lim 1
■ 17. If 2 x x2 16. lim
■ ■ ■ 2 x for 0 fx 18. Prove that lim x l 0 x cos 1 x
 27 x l5 21. lim x 2 sx 2
x2 ■ x 1 ■ ■ ■ 20. lim sx s2x
2x
■ ■ ■ ■ sx xl4 ■ ■ ■ if x
if 0
if x 2 ■ 4
■ 0
x
3 (a) Evaluate each limit, if it exists.
(i) lim f x
(ii) lim f x
x l0 ■ ■ ■ 3 ■ ■ (v) lim f x ■ 9 2x 2 30. Find equations of the tangent lines to the curve 2
1 3x at the points with xcoordinates 0 and 1. line is given by s 1 2 t t 2 4, where t is measured in
seconds.
(a) Find the average velocity over each time period.
(i) 1, 3
(ii) 1, 2
(iii) 1, 1.5
(iv) 1, 1.1
(b) Find the instantaneous velocity when t 1. is held ﬁxed, then the product of the pressure P and the volume
V is a constant. Suppose that, for a certain gas, PV 800,
where P is measured in pounds per square inch and V is measured in cubic inches.
(a) Find the average rate of change of P as V increases from
200 in3 to 250 in3.
(b) Express V as a function of P and show that the instantaneous rate of change of V with respect to P is inversely proportional to the square of P. x l0 (vi) lim f x x l3 ■ 32. According to Boyle’s Law, if the temperature of a conﬁned gas (iii) lim f x x l0 (iv) lim f x ■ 31. The displacement (in meters) of an object moving in a straight 23. Let x l3 0, 1
■ at the point 2, 1 .
(b) Find an equation of this tangent line. 2 22. lim sx
3x
x3 2 x,
■ 1 ) 0 xl0 fx 3
■ 2, 0. 3 ■ 0, y 2 2 ■ 2 3, ﬁnd lim x l1 f x . x 8 3x xl2
■ 28. 2 sin x Prove the statement using the precise deﬁnition of a limit. 19. lim 7x ■ 9 x2 29. (a) Find the slope of the tangent line to the curve y x l2 2 19–22 27. 2 x 3 2v 8
4
v
16 xl9 ■ 1 v 14. lim (s x 123 27–28  Use the Intermediate Value Theorem to show that there is
a root of the equation in the given interval. v ■ s1
x ■ v2 v l2 8
8 x l0 ■ vl4 ss
16
x
x 13. lim 4
4 10. lim 4 ❙❙❙❙ x l3 (b) Where is f discontinuous?
(c) Sketch the graph of f .
24. Let 2x x 2
2x
x4 tx if
if
if
if 0
2
3
x x
x
x
4 ; 33. Use a graph to ﬁnd a number such that 2
3
4 x
x (a) For each of the numbers 2, 3, and 4, discover whether t is
continuous from the left, continuous from the right, or continuous at the number.
(b) Sketch the graph of t.
25–26 Show that the function is continuous on its domain. State
the domain.
 25. h x 4
sx 26. t x sx 2
x2 ■ ■ ■ 0.2 whenever x y x
x 1
1 and the tangent lines to this curve at the points 2, 3
and 1, 0 . 36. Let f x
■ ■ ■ ■ ■ ■ ■ 2 ; 34. Graph the curve t x for all x, where lim x l a t x Find lim x l a f x . 9
2
■ 3 35. Suppose that f x x 3 cos x ■ 1
1 x
x.
(a) For what values of a does lim x l a f x exist?
(b) At what numbers is f discontinuous? 0. 5E02(pp 124125) 1/17/06 12:21 PM PROBLEMS
PLUS Page 124 In our discussion of the principles of problem solving we considered the problemsolving
strategy of introducing something extra (see page 58). In the following example we show
how this principle is sometimes useful when we evaluate limits. The idea is to change the
variable—to introduce a new variable that is related to the original variable—in such a way
as to make the problem simpler. Later, in Section 5.5, we will make more extensive use of
this general idea.
EXAMPLE 1 Evaluate lim 3
s1 cx
x xl0 1 , where c is a constant. SOLUTION As it stands, this limit looks challenging. In Section 2.3 we evaluated several limits in which both numerator and denominator approached 0. There our strategy was to perform some sort of algebraic manipulation that led to a simplifying cancellation, but here
it’s not clear what kind of algebra is necessary.
So we introduce a new variable t by the equation
3
s1 t cx We also need to express x in terms of t, so we solve this equation:
t3 1 cx t3 x 1
c Notice that x l 0 is equivalent to t l 1. This allows us to convert the given limit into one
involving the variable t :
3
s1 cx
x xl0 lim 1 t
t3 lim lim ct
t3 t l1 t l1 1
1c
1
1 The change of variable allowed us to replace a relatively complicated limit by a simpler
one of a type that we have seen before. Factoring the denominator as a difference of
cubes, we get
lim
t l1 ct
t3 1
1 lim
t l1 lim
t l1 t
t2 ct 1
1 t2 t
c
t 1 1
c
3 The following problems are meant to test and challenge your problemsolving skills.
Some of them require a considerable amount of time to think through, so don’t be discouraged if you can’t solve them right away. If you get stuck, you might ﬁnd it helpful to refer
to the discussion of the principles of problem solving on page 58. 124 5E02(pp 124125) 1/17/06 12:22 PM P RO B L E M S Page 125 3
sx
sx 1. Evaluate lim
x l1 1
.
1 2. Find numbers a and b such that lim sax 2x 3. Evaluate lim 1 2x 1 x xl0 b 2 1. x xl0 . x 2 and the point Q where the perpendicular
bisector of OP intersects the yaxis. As P approaches the origin along the parabola, what
happens to Q ? Does it have a limiting position? If so, ﬁnd it. 4. The ﬁgure shows a point P on the parabola y y y=≈
Q P 5. If x denotes the greatest integer function, ﬁnd lim x l x x .
6. Sketch the region in the plane deﬁned by each of the following equations. (a) x
0 x 2 y 2 1 (b) x 2 y 2 (c) x 3 7. Find all values of a such that f is continuous on FIGURE FOR PROBLEM 4 fx x
x2 y 2 1 (d) x y 1 : 1 if x
if x a
a 8. A ﬁxed point of a function f is a number c in its domain such that f c c. (The function
doesn’t move c; it stays ﬁxed.)
(a) Sketch the graph of a continuous function with domain 0, 1 whose range also lies
in 0, 1 . Locate a ﬁxed point of f .
(b) Try to draw the graph of a continuous function with domain 0, 1 and range in 0, 1 that
does not have a ﬁxed point. What is the obstacle?
(c) Use the Intermediate Value Theorem to prove that any continuous function with domain
0, 1 and range a subset of 0, 1 must have a ﬁxed point. 9. If lim x l a f x
A 2 and lim x l a f x tx 1, ﬁnd lim x l a f x t x . 10. (a) The ﬁgure shows an isosceles triangle ABC with B
C. The bisector of angle B
intersects the side AC at the point P. Suppose that the base BC remains ﬁxed but the
altitude AM of the triangle approaches 0, so A approaches the midpoint M of BC. What
happens to P during this process? Does it have a limiting position? If so, ﬁnd it.
(b) Try to sketch the path traced out by P during this process. Then ﬁnd the equation of this
curve and use this equation to sketch the curve. P B tx M FIGURE FOR PROBLEM 10 C 11. (a) If we start from 0 latitude and proceed in a westerly direction, we can let T x denote the temperature at the point x at any given time. Assuming that T is a continuous function
of x, show that at any ﬁxed time there are at least two diametrically opposite points on the
equator that have exactly the same temperature.
(b) Does the result in part (a) hold for points lying on any circle on Earth’s surface?
(c) Does the result in part (a) hold for barometric pressure and for altitude above sea level? 125 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas.
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