580
● Chapter 9
OBJECTIVES
●
●
●
Derivatives
9.1
To use graphs and numerical
tables to find limits of
functions, when they exist
To find limits of polynomial
functions
To find limits of rational
functions
Limits
] Application Preview
Although everyone recognizes the value of eliminating any and all particulate pollution from
smokestack emissions of factories, company owners are concerned about the cost of removing this
pollution. Suppose that USA Steel has shown that the cost C of removing p percent of the particulate pollution from the emissions at one of its plants is
C ! C(p) !
7300p
100 " p
To investigate the cost of removing as much of the pollution as possible, we can evaluate the limit
as p (the percent) approaches 100 from values less than 100. (See Example 6.) Using a limit is
important in this case, because this function is undefined at p ! 100 (it is impossible to remove
100% of the pollution).
In various applications we have seen the importance of the slope of a line as a rate of
change. In particular, the slope of a linear total cost, total revenue, or profit function for a
product tells us the marginals or rates of change of these functions. When these functions
are not linear, how do we define marginals (and slope)?
We can get an idea about how to extend the notion of slope (and rate of change) to
functions that are not linear. Observe that for many curves, if we take a very close (or
“zoom-in”) view near a point, the curve appears straight. See Figure 9.1. We can think of
the slope of the “straight” line as the slope of the curve. The mathematical process used to
obtain this “zoom-in” view is the process of taking limits.
y
y = f (x)
P
x
P
P
Figure 9.1
Notion of a Limit
Zooming in near point P, the curve appears straight.
We have used the notation f(c) to indicate the value of a function f(x) at x ! c. If we need
to discuss a value that f(x) approaches as x approaches c, we use the idea of a limit. For
example, if
f(x) !
x2 # x # 6
x$2
then we know that x ! #2 is not in the domain of f(x), so f(#2) does not exist even though
f(x) exists for every value of x " #2. Figure 9.2 shows the graph of y ! f(x) with an open
9.1 Limits ●
581
y
x
−2
2
−2
f (x) =
−4
x2 − x − 6
x+2
(−2, −5)
Figure 9.2
TABLE 9.1
Left of "2
x
#3.000
#2.500
#2.100
#2.010
#2.001
f(x) !
x2 " x " 6
x#2
#6.000
#5.500
#5.100
#5.010
#5.001
Right of "2
x
#1.000
#1.500
#1.900
#1.990
#1.999
f(x) !
x2 " x " 6
x#2
#4.000
#4.500
#4.900
#4.990
#4.999
Limit
−6
circle where x ! #2. The open circle indicates that f(#2) does not exist but shows that
points near x ! #2 have functional values that lie on the line on either side of the open
circle. Even though f(#2) is not defined, the figure shows that as x approaches #2 from
either side of #2, the graph approaches the open circle at (#2, #5) and the values of f(x)
approach #5. Thus #5 is the limit of f(x) as x approaches #2, and we write
lim f(x) ! #5, or
xS#2
f(x) S #5 as x S #2
This conclusion is fairly obvious from the graph, but it is not so obvious from the equation
for f(x).
We can use the values near x ! #2 in Table 9.1 to help verify that f(x) S #5 as
x S #2. Note that to the left of #2, the values of x increase from #3.000 to #2.001 in
small increments, and in the corresponding column for f(x), the values of the function f(x)
increase from #6.000 to #5.001. To the right of #2, the values of x decrease from #1.000
to #1.999 while the corresponding values of f(x) decrease from #4.000 to #4.999. Hence,
Table 9.1 and Figure 9.2 indicate that the value of f(x) approaches #5 as x approaches #2
from both sides of x ! #2.
From our discussion of the graph in Figure 9.2 and Table 9.1, we see that as x
approaches #2 from either side of #2, the limit of the function is the value L that the function approaches. This limit L is not necessarily the value of the function at x ! #2. This
leads to our intuitive definition of limit.
Let f(x) be a function defined on an open interval containing c, except perhaps at c.
Then
lim f(x) ! L
xSc
is read “the limit of f(x) as x approaches c equals L.” The number L exists if we can
make values of f(x) as close to L as we desire by choosing values of x sufficiently close
to c. When the values of f(x) do not approach a single finite value L as x approaches c,
we say the limit does not exist.
As the definition states, a limit as x S c can exist only if the function approaches a
single finite value as x approaches c from both the left and right of c.
582
● Chapter 9
Derivatives
● EXAMPLE 1 Limits
Figure 9.3 shows three functions for which the limit exists as x approaches 2. Use this figure
to find the following.
(a) lim f(x) and f(2) (if it exists)
xS2
(b) lim g(x) and g(2) (if it exists)
xS2
(c) lim h(x) and h(2) (if it exists)
xS2
y
y
y
4
4
4
(2, 4)
(2, 3)
2
2
−2
x
2
−1
4
2
y = g(x)
y = f (x)
x
−2
1
4
(2, −1)
−2
Figure 9.3
(a)
(b)
(2, 1)
x
−2
2
4
y = h(x)
−2
(c)
Solution
(a) From the graph in Figure 9.3(a), we see that as x approaches 2 from both the left
and the right, the graph approaches the point (2, 3). Thus f(x) approaches the single
value 3. That is,
lim f(x) ! 3
xS2
The value of f(2) is the y-coordinate of the point on the graph at x ! 2. Thus f(2) ! 3.
(b) Figure 9.3(b) shows that as x approaches 2 from both the left and the right, the graph
approaches the open circle at (2, #1). Thus
lim g(x) ! #1
xS2
The figure also shows that at x ! 2 there is no point on the graph. Thus g(2) is
undefined.
(c) Figure 9.3(c) shows that
lim h(x) ! 1
xS2
The figure also shows that at x ! 2 there is a point on the graph at (2, 4). Thus
h(2) ! 4, and we see that lim h(x) " h(2).
xS2
As Example 1 shows, the limit of the function as x approaches c may or may not be
the same as the value of the function at x ! c.
In Example 1 we saw that the limit as x approaches 2 meant the limit as x approaches
2 from both the left and the right. We can also consider limits only from the left or only
from the right; these are called one-sided limits.
583
9.1 Limits ●
Limit from the Right:
One-Sided Limits
lim f(x) ! L
xSc$
means the values of f(x) approach the value L as x S c but x ' c.
Limit from the Left:
lim f(x) ! M
xSc#
means the values of f(x) approach the value M as x S c but x & c.
Note that when one or both one-sided limits fail to exist, then the limit does not
exist. Also, when the one-sided limits differ, such as if L " M above, then the values of
f(x) do not approach a single value as x approaches c, and lim f(x) does not exist. So far,
xSc
the examples have illustrated limits that existed and for which the two one-sided limits were
the same. In the next example, we use one-sided limits and consider cases where a limit
does not exist.
● EXAMPLE 2 One-Sided Limits
Using the functions graphed in Figure 9.4, determine why the limit as x S 2 does not
exist for
(a) f(x)
(b) g(x)
(c) h(x)
4 − x 2 if x ≤ 2
2x − 1 if x > 2
h(x) =
y
y
y
6
f (x) =
5
7
3
1
(x − 2) 2
6
2
4
5
4
1
3
2
2
-1
1
1
2
(a)
3
4
3
4
1
g(x) =
x−2
-2
x
Figure 9.4
3
x
1
-3
(b)
2
1
x
-1
-1
1
2
3
4
(c)
Solution
(a) As x S 2 from the left side and the right side of x ! 2, f(x) increases without bound,
which we denote by saying that f(x) approaches $% as x S 2. In this case, lim f(x)
xS2
does not exist [denoted by lim f(x) DNE] because f(x) does not approach a finite
xS2
value as x S 2. In this case, we write
f(x) S $% as x S 2
The graph has a vertical asymptote at x ! 2.
(b) As x S 2 from the left, g(x) approaches #%, and as x S 2 from the right, g(x)
approaches $%, so g(x) does not approach a finite value as x S 2. Therefore, the
limit does not exist. The graph of y ! g(x) has a vertical asymptote at x ! 2.
584
● Chapter 9
Derivatives
In this case we summarize by writing
lim g(x) DNE
or
g(x) S #% as x S 2#
lim g(x) DNE
or
g(x) S $% as x S 2$
xS2#
xS2$
and
lim g(x) DNE
xS2
(c) As x S 2 from the left, the graph approaches the point at (2, 0), so lim# h(x) ! 0. As
xS2
x S 2 from the right, the graph approaches the open circle at (2, 3), so lim$ h(x) ! 3.
xS2
Because these one-sided limits differ, lim h(x) does not exist.
xS2
Examples 1 and 2 illustrate the following two important facts regarding limits.
1. The limit of a function as x approaches c is independent of the value of the
function at c. When lim f(x) exists, the value of the function at c may be (i) the
xSc
same as the limit, (ii) undefined, or (iii) defined but different from the limit (see
Figure 9.3 and Example 1).
2. The limit is said to exist only if the following conditions are satisfied:
(a) The limit L is a finite value (real number).
(b) The limit as x approaches c from the left equals the limit as x approaches c
from the right. That is, we must have
lim f(x) ! lim$ f(x)
xSc#
xSc
Figure 9.4 and Example 2 illustrate cases where lim f(x) does not exist.
xSc
●
Checkpoint
Properties of Limits,
Algebraic Evaluation
1.
2.
3.
4.
Can lim# f(x) exist if f (c) is undefined?
xSc
Does lim f(x) exist if f(c) ! 0?
xSc
Does f(c) ! 1 if lim f(x) ! 1?
xSc
If lim# f(x) ! 0, does lim f(x) exist?
xSc
xSc
Fact 1 regarding limits tells us that the value of the limit of a function as x S c will not
always be the same as the value of the function at x ! c. However, there are many functions for which the limit and the functional value agree [see Figure 9.3(a)], and for these
functions we can easily evaluate limits. The following properties of limits allow us to identify certain classes or types of functions for which lim f(x) equals f(c).
xSc
Properties of Limits
If k is a constant, lim f(x) ! L, and lim g(x) ! M, then the following are true.
xSc
I. lim k ! k
xSc
II. lim x ! c
xSc
III. lim 3 f(x) ( g(x)4 ! L ( M
xSc
xSc
IV. lim 3 f(x) $ g(x)4 ! LM
xSc
V. lim
xSc
f(x)
L
!
g(x)
M
n
if M " 0
n
n
VI. lim 1 f(x) ! 1 lim f (x) ! 1 L,
xSc
xSc
provided that L ' 0 when n is even.
9.1 Limits ●
585
lim f(x) can be found
If f is a polynomial function, then Properties I–IV imply that xSc
by evaluating f(c). Moreover, if h is a rational function whose denominator is not zero at
x ! c, then Property V implies that lim h(x) can be found by evaluating h(c). The followxSc
ing summarizes these observations and recalls the definitions of polynomial and rational
functions.
Limit
Function
Definition
Polynomial function
The function
f(x) ! anxn $ an#1xn#1 $ p $ a1x $ a0,
where an " 0 and n is a positive integer, is called a
polynomial function of degree n.
Rational function
lim f(x) ! f (c)
xSc
for all values c (by Properties I–IV)
The function
f(x)
h(x) !
g(x)
where both f(x) and g(x) are polynomial functions, is
called a rational function.
f(x)
f (c)
!
g(x)
g(c)
when g(c) " 0 (by Property V)
lim h(x) ! lim
xSc
xSc
● EXAMPLE 3 Limits
Find the following limits, if they exist.
(a) lim (x3 # 2x)
xS#1
(b) lim
xS4
x2 # 4x
x#2
Solution
(a) Note that f (x) ! x3 # 2x is a polynomial, so
lim f(x) ! f (#1) ! (#1)3 # 2(#1) ! 1
xS#1
Figure 9.5(a) shows the graph of f(x) ! x3 # 2x.
(b) Note that this limit has the form
lim
xSc
f(x)
g(x)
where f(x) and g(x) are polynomials and g(c) " 0. Therefore, we have
lim
xS4
42 # 4(4)
0
x2 # 4x
!
! !0
x#2
4#2
2
Figure 9.5(b) shows the graph of g(x) !
x2 # 4x
.
x#2
586
● Chapter 9
Derivatives
y
y
4
4
2
2
g(x) =
x 2 − 4x
x−2
(−1, 1)
(4, 0)
x
x
−2
2
−2
f (x) = x 3 − 2x
−2
4
6
−2
−4
Figure 9.5
2
-4
(b)
(a)
We have seen that we can use Property V to find the limit of a rational function
f(x)!g(x) as long as the denominator is not zero. If the limit of the denominator of f(x)!g(x)
is zero, then there are two possible cases.
I. Both lim g(x) ! 0 and lim f(x) ! 0, or
xSc
xSc
II. lim g(x) ! 0 and lim f(x) " 0.
xSc
xSc
In case I we say that f(x)!g(x) has the form 0!0 at x ! c. We call this the 0!0 indeterminate form; the limit cannot be evaluated until x # c is factored from both f(x) and
g(x) and the fraction is reduced. Example 4 will illustrate this case.
In case II, the limit has the form a!0, where a is a constant, a " 0. This expression is
undefined, and the limit does not exist. Example 5 will illustrate this case.
● EXAMPLE 4 0!0 Indeterminate Form
Evaluate the following limits, if they exist.
x2 # 4
x#2
x2 # 3x $ 2
(b) lim
xS1
x2 # 1
(a) lim
xS2
Solution
(a) We cannot find the limit by using Property V because the denominator is zero at
x ! 2. The numerator is also zero at x ! 2, so the expression
x2 # 4
x#2
has the 0!0 indeterminate form at x ! 2. Thus we can factor x # 2 from both the
numerator and the denominator and reduce the fraction. (We can divide by x # 2
because x # 2 " 0 while x S 2.)
lim
xS2
(x # 2)(x $ 2)
x2 # 4
! lim
! lim (x $ 2) ! 4
xS2
xS2
x#2
x#2
9.1 Limits ●
587
Figure 9.6(a) shows the graph of f(x) ! (x2 # 4)!(x # 2). Note the open circle
at (2, 4).
(b) By substituting 1 for x in (x2 # 3x $ 2)!(x2 # 1), we see that the expression has the
0!0 indeterminate form at x ! 1, so x # 1 is a factor of both the numerator and the
denominator. (We can then reduce the fraction because x # 1 " 0 while x S 1.)
lim
xS1
(x # 1)(x # 2)
x2 # 3x $ 2
! lim
2
xS1
x #1
(x # 1)(x $ 1)
x#2
! lim
xS1 x $ 1
1#2
#1
!
!
(by Property V)
1$1
2
Figure 9.6(b) shows the graph of g(x) ! (x2 # 3x $ 2)!(x2 # 1). Note the open
circle at (1, #12).
y
y
6
4
4
(2, 4)
2
2
f (x) =
x
x2 − 4
x−2
-2
x
−2
Figure 9.6
2
x 2 − 3x + 2
x2 − 1
g(x) =
4
(a)
2
(1, )
– 12
4
-2
(b)
Note that although both problems in Example 4 had the 0!0 indeterminate form, they
had different answers.
● EXAMPLE 5 Limit with a!0 Form
Find lim
xS1
x2 $ 3x $ 2
, if it exists.
x#1
Solution
Substituting 1 for x in the function results in 6!0, so this limit has the form a!0, with a " 0,
and is like case II discussed previously. Hence the limit does not exist. Because the numerator is not zero when x ! 1, we know that x # 1 is not a factor of the numerator, and we
cannot divide numerator and denominator as we did in Example 4. Table 9.2 confirms that
this limit does not exist, because the values of the expression are unbounded near x ! 1.
588
● Chapter 9
Derivatives
TABLE 9.2
Left of x ! 1
Right of x ! 1
x2 # 3x # 2
x"1
x
x2 # 3x # 2
x"1
x
0
#2
0.5
#7.5
0.7
#15.3
0.9
#55.1
0.99
#595.01
0.999
#5,995.001
0.9999
#59,995.0001
x2 $ 3x $ 2
lim
DNE
xS1#
x#1
(f (x) S #% as x S 1#)
2
12
1.5
17.5
1.2
35.2
1.1
65.1
1.01
605.01
1.001
6,005.001
1.0001 60,005.0001
x2 $ 3x $ 2
lim
DNE
xS1$
x#1
(f (x) S $% as x S 1$)
The left-hand and right-hand limits do not exist. Thus lim
xS1
x2 $ 3x $ 2
does not exist.
x#1
In Example 5, even though the left-hand and right-hand limits do not exist (see Table
9.2), knowledge that the functional values are unbounded (that is, that they become infinite)
is helpful in graphing. The graph is shown in Figure 9.7. We see that x ! 1 is a vertical
asymptote.
f (x) unbounded ( f (x) → + %)
as x → 1+
y
20
16
12
8
y=
4
x 2 + 3x + 2
x−1
x
−12 −8
Figure 9.7
4
−4
8
12
16
f (x) unbounded ( f (x) → − %)
as x → 1−
−8
The results of Examples 4 and 5 can be summarized as follows.
Rational Functions:
Evaluating Limits of the
f (x)
Form lim
where
xSc g(x)
lim g(x) ! 0
xSc
Type I. If lim f(x) ! 0 and lim g(x) ! 0, then the fractional expression has the 0!0
xSc
xSc
indeterminate form at x ! c. We can factor x # c from f(x) and g(x), reduce the
fraction, and then find the limit of the resulting expression, if it exists.
f(x)
does not exist. In this case,
g(x)
the values of f(x)!g(x) are unbounded near x ! c; the line x ! c is a vertical asymptote.
Type II. If lim f(x) " 0 and lim g(x) ! 0, then lim
xSc
xSc
xSc
9.1 Limits ●
589
] EXAMPLE 6 Cost-Benefit (Application Preview)
USA Steel has shown that the cost C of removing p percent of the particulate pollution
from the smokestack emissions at one of its plants is
C ! C(p) !
7300p
100 # p
To investigate the cost of removing as much of the pollution as possible, find:
(a)
(b)
(c)
(d)
the
the
the
the
cost
cost
cost
cost
of
of
of
of
removing
removing
removing
removing
50% of the pollution.
90% of the pollution.
99% of the pollution.
100% of the pollution.
Solution
(a) The cost of removing 50% of the pollution is $7300 because
C(50) !
7300(50)
365,000
!
! 7300
100 # 50
50
(b) The cost of removing 90% of the pollution is $65,700 because
C(90) !
7300(90)
657,000
!
! 65,700
100 # 90
10
(c) The cost of removing 99% of the pollution is $722,700 because
C(99) !
7300(99)
722,700
!
! 722,700
100 # 99
1
(d) The cost of removing 100% of the pollution is undefined because the denominator of
the function is 0 when p ! 100. To see what the cost approaches as p approaches
7300p
. This limit has the
100 from values smaller than 100, we evaluate lim #
xS100 100 # p
7300
S $% as x S 100#, which means
Type II form for rational functions. Thus
100 # p
that as the amount of pollution that is removed approaches 100%, the cost increases
without bound. (That is, it is impossible to remove 100% of the pollution.)
●
Checkpoint
5. Evaluate the following limits (if they exist).
2x2 $ 5x # 3
x2 # 3x # 3
(a) lim
(b) lim 2
(c)
2
xS#3
xS5 x # 8x $ 1
x #9
In Problems 6–9, assume that f, g, and h are polynomials.
lim
xS#3!4
4x
4x $ 3
6. Does lim f(x) ! f (c)?
xSc
7. Does lim
xSc
g(x)
g(c)
!
?
h(x)
h(c)
8. If g(c) ! 0 and h(c) ! 0, can we be certain that
g(x)
g(x)
! 0?
(a) lim
(b) lim
exists?
xSc h(x)
xSc h(x)
9. If g(c) " 0 and h(c) ! 0, what can be said about lim
xSc
g(x)
h(x)
and lim
?
xSc g(x)
h(x)
590
● Chapter 9
Derivatives
Limits of Piecewise
Defined Functions
As we noted in Section 2.4, “Special Functions and Their Graphs,” many applications are
modeled by piecewise defined functions. To see how we evaluate a limit involving a piecewise defined function, consider the following example.
● EXAMPLE 7 Limit of a Piecewise Defined Function
Find lim# f(x), lim$ f (x), and lim f(x), if they exist, for
xS1
TABLE 9.3
xS1
xS1
f(x) ! b
Left of 1
x
f(x) ! x2 # 1
0.1
0.9
0.99
0.999
0.9999
1.01
1.81
1.98
1.998
1.9998
Solution
Because f(x) is defined by x2 $ 1 when x & 1,
lim f(x) ! lim# (x2 $ 1) ! 2
xS1#
f(x) ! x # 2
1.2
1.01
1.001
1.0001
1.00001
3.2
3.01
3.001
3.0001
3.00001
xS1
Because f(x) is defined by x $ 2 when x ' 1,
Right of 1
x
x2 $ 1 for x ) 1
x $ 2 for x ' 1
lim f(x) ! lim$(x $ 2) ! 3
xS1$
xS1
And because
2 ! lim# f(x) " lim$ f(x) ! 3
xS1
xS1
lim f(x) does not exist.
Table 9.3 and Figure 9.8 show these results numerically and graphically.
xS1
y
f (x) = x + 2
(x > 1)
6
5
4
3
2
f (x) = x 2 + 1
(x ≤ 1)
1
Figure 9.8
Calculator
Note
x
-2
-1
0
1
2
3
4
We have used graphical, numerical, and algebraic methods to understand and evaluate limits. Graphing calculators can be especially effective when we are exploring limits graphi■
cally or numerically.
● EXAMPLE 8 Limits: Graphically, Numerically, and Algebraically
Consider the following limits.
(a) lim
xS5
x2 $ 2x # 35
x2 # 6x $ 5
(b) lim
xS#1
2x
x$1
Investigate each limit by using the following methods.
(i) Graphically: Graph the function with a graphing utility and trace near the limiting
x-value.
(ii) Numerically: Use the table feature of a graphing utility to evaluate the function very
close to the limiting x-value.
(iii) Algebraically: Use properties of limits and algebraic techniques.
9.1 Limits ●
591
Solution
x2 $ 2x # 35
(a) lim 2
xS5 x # 6x $ 5
(i) Figure 9.9(a) shows the graph of y ! (x2 $ 2x # 35)!(x2 # 6x $ 5). Tracing
near x ! 5 shows y-values getting close to 3.
(ii) Figure 9.9(b) shows a table for y1 ! (x2 $ 2x # 35)!(x2 # 6x $ 5) with
x-values approaching 5 from both sides (note that the function is undefined
at x ! 5). Again, the y-values approach 3.
x2 $ 2x # 35
! 3.
Both (i) and (ii) strongly suggest lim 2
xS5 x # 6x $ 5
(iii) Algebraic evaluation of this limit confirms what the graph and the table suggest.
lim
xS5
(x $ 7)(x # 5)
12
x2 $ 2x # 35
x$7
! lim
! lim
!
!3
2
xS5
xS5
x # 6x $ 5
(x # 1)(x # 5)
x#1
4
12
1
-4.5
14.5
X=4.8
X
Y1
4.9
4.99
4.999
5
5.001
5.01
5.1
3.0513
3.005
3.0005
ERROR
2.9995
2.995
2.9512
X=4.9
Y=3.1052632
-6
(a)
Figure 9.9
(b)
2x
x$1
(i) Figure 9.10(a) shows the graph of y ! 2x!(x $ 1); it indicates a break in the
graph near x ! #1. Evaluation confirms that the break occurs at x ! #1 and
also suggests that the function becomes unbounded near x ! #1. In addition, we
can see that as x approaches #1 from opposite sides, the function is headed in
different directions. All this suggests that the limit does not exist.
(ii) Figure 9.10(b) shows a graphing calculator table of values for y1 ! 2x!(x $ 1)
and with x-values approaching x ! #1. The table reinforces our preliminary
conclusions from the graph that the limit does not exist, because the function is
unbounded near x ! #1.
2x
DNE.
(iii) Algebraically we see that this limit has the form #2!0. Thus lim
xS#1 x $ 1
(b) lim
xS#1
45
1
-3.4
X
1.4
X=-1.05
-1.1
-1.01
-1.001
-1
-.999
-.99
-.9
Y1
22
202
2002
ERROR
-1998
-198
-18
X=-.9
Y=42
-45
Figure 9.10
(a)
(b)
We could also use the graphing and table features of spreadsheets to explore limits.
592
●
● Chapter 9
Derivatives
Checkpoint Solutions
1. Yes. For example, Figure 9.2 and Table 9.1 show that this is possible for
x2 # x # 6
f(x) !
. Remember that lim f(x) does not depend on f(c).
xSc
x$2
2. Not necessarily. Figure 9.4(c) shows the graph of y ! h(x) with h(2) ! 0, but lim h(x)
xS2
does not exist.
3. Not necessarily. Figure 9.3(c) shows the graph of y ! h(x) with lim h(x) ! 1 but
xS2
h(2) ! 4.
4. Not necessarily. For example, Figure 9.4(c) shows the graph of y ! h(x) with
lim h(x) ! 0, but with lim$ h(x) ! 2, so the limit doesn’t exist. Recall that if
xS2#
xS2
lim f(x) ! lim$ f(x) ! L, then lim f(x) ! L.
xSc#
xSc
xSc
(2x # 1)(x $ 3)
2x2 $ 5x # 3
2x # 1
#7
7
! lim
! lim
!
!
5. (a) lim
xS#3
xS#3 (x $ 3)(x # 3)
xS#3 x # 3
x2 # 9
#6
6
x2 # 3x # 3
7
1
!
!#
(b) lim 2
xS5 x # 8x $ 1
#14
2
4x
(c) Substituting x ! #3!4 gives #3!0, so lim
does not exist.
xS#3!4 4x $ 3
6. Yes, Properties I–IV yield this result.
7. Not necessarily. If h(c) " 0, then this is true. Otherwise, it is not true.
8. For both (a) and (b), g(x)!h(x) has the 0!0 indeterminate form at x ! c. In this case we
can make no general conclusion about the limit. It is possible for the limit to exist (and
be zero or nonzero) or not to exist. Consider the following 0!0 indeterminate forms.
x(x $ 1)
x2
! lim x ! 0
! lim (x $ 1) ! 1
(i) lim
(ii) lim
xS0 x
xS0
xS0
xS0
x
x
1
(iii) lim 2 ! lim , which does not exist
xS0 x
xS0 x
g(x)
h(x)
9. lim
does not exist and lim
!0
xSc h(x)
xSc g(x)
9.1 Exercises
In Problems 1–6, a graph of y ! f(x) is shown and a
c-value is given. For each problem, use the graph to find
the following, whenever they exist.
(a) lim f(x) and (b) f(c)
xSc
1. c ! 4
2. c ! 6
4. c ! #10
y
20
y
y = f (x)
9
8
4
–4
10
y = f (x)
x
6. c ! #2
y
3
–3
5
– 1 5 – 10 – 5
5. c ! #8
x
y = f (x)
x
20
−10
3
y = f (x)
15
10
y
y
–8
3. c ! 20
9
x
y = f (x)
−12 −8
−4
y
4
2
x
−4
−8
−6
−2
2
−2
y = f (x)
x
9.1 Limits ●
In Problems 7–10, use the graph of y ! f(x) and the
given c-value to find the following, whenever they exist.
(a) lim" f(x)
(b) lim# f(x)
xSc
xSc
(c) lim f(x)
(d) f(c)
xSc
8. c ! 2
7. c ! #10
y
y
y = f (x)
-
4
10
y = f (x)
2
–20
x
–10
−2
x
2
−2
y
3
–9
–6
3
x
1
−3
−1
−6
−2
−3
2
x
3
y = f (x)
In Problems 11–14, complete each table and predict the
limit, if it exists.
2x $ 1
2 # x # x2
11. f (x) !
12. f(x) ! 1
2
x#1
4 # x
lim f(x) ! ?
lim f (x) ! ?
xS1
x
0.9
0.99
0.999
T
1
c
1.001
1.01
1.1
xS#0.5
f(x)
T
T
?
c
c
#0.4999
#0.499
#0.49
f(x)
T
T
c
c
1.001
1.01
1.1
T
c
xS1
1
f(x)
#0.5
5x # 1
8 # 2x # x2
lim f(x) ! ?
x
0.9
0.99
0.999
x
#0.51
#0.501
#0.5001
?
13. f(x) ! b
?
for x & 1
for x * 1
f(x)
T
#2
?
c
c
xS#35
1
–3
y = f (x)
x
#2.1
#2.01
#2.001
In Problems 15–36, use properties of limits and algebraic
methods to find the limits, if they exist.
15. lim (34 $ x)
16. lim (82 # x)
10. c ! 2
y
xS#2
#1.999
#1.99
−4
9. c ! #412
for x ) #2
for x ' #2
14. f (x) ! b
T
4
5
4 # x2
x2 $ 2x
lim f(x) ! ?
593
xS80
17. lim (4x3 # 2x2 $ 2)
xS#1
18. lim (2x3 # 12x2 $ 5x $ 3)
xS3
4x # 2
1 # 3x
19. lim
20. lim
xS#1!2 4x2 $ 1
xS#1!3 9x2 $ 1
x2 # 9
x2 # 16
21. lim
22. lim
xS3 x # 3
xS#4 x $ 4
x2 # 8x $ 7
x2 $ 8x $ 15
23. lim 2
24. lim
xS7 x # 6x # 7
xS#5
x2 $ 5x
x2 $ 4x $ 4
x2 # 8x # 20
25. lim 2
26. lim 2
xS#2 x $ 3x $ 2
xS10 x # 11x $ 10
10 # 2x for x & 3
27. lim f(x), where f(x) ! b 2
xS3
for x * 3
x #x
7x # 10 for x & 5
28. lim f(x), where f(x) ! b
xS5
25 for x * 5
4
x2 $
for x ) #1
x
c
29. lim f(x), where f (x) !
xS#1
3x3 # x # 1 for x ' #1
x3 # 4
for x ) 2
x#3
30. lim f(x), where f(x) ! d
xS2
3 # x2
for x ' 2
x
2
x $ 6x $ 9
x2 # 6x $ 8
31. lim
32. lim
xS2
xS5
x#2
x#5
x2 $ 5x $ 6
x2 $ 2x # 3
33. lim
34. lim
xS#1
xS3
x$1
x#3
3
3
(x $ h) # x
2(x $ h)2 # 2x2
35. lim
36. lim
hS0
hS0
h
h
In Problems 37–40, graph each function with a graphing
utility and use it to predict the limit. Check your work
either by using the table feature of the graphing utility
or by finding the limit algebraically.
x2 # 19x $ 90
x4 $ 3x3
37. lim
38. lim
2
xS10
xS#3 2x4 # 18x2
3x # 30x
594
● Chapter 9
39. lim
xS#1 x2
Derivatives
x3 # x
$ 2x $ 1
40. lim
xS5
x2 # 7x $ 10
x2 # 10x $ 25
In Problems 41–44, use the table feature of a graphing
utility to predict each limit. Check your work by using
either a graphical or an algebraic approach.
x4 # 4x2
x3 $ 4x2
41. lim 2
42. lim
2
xS#2 x $ 8x $ 12
xS#4 2x $ 7x # 4
3
12 # x for x ) 4
4
c
43. lim f(x), where f(x) !
xS4
x2 # 7
for x ' 4
2 $ x # x2 for x ) 7
44. lim f(x), where f(x) ! b
xS7
13 # 9x
for x ' 7
(a) Find lim$ S(x).
S ! S(x) !
to three decimal places. This limit equals the special
number e that is discussed in Section 5.1, “Exponential
Functions,” and Section 6.2, “Compound Interest; Geometric Sequences.”
46. If lim 3 f(x) $ g(x)4 ! 5 and lim g(x) ! 11, find
xS2
xS2
(a) lim f(x)
xS2
(b) lim 5 3 f(x)4 2 # 3g(x)4 2 6
xS2
3g(x)
(c) lim
xS2 f(x) # g(x)
47. If lim f(x) ! 4 and lim g(x) ! #2, find
xS3
xS3
(a) lim 3 f(x) $ g(x)4
(b) lim 3 f(x) # g(x)4
xS3
xS3
g(x)
(c) lim 3 f(x) $ g(x)4
(d) lim
xS3
xS3 f(x)
48. (a) If lim$ f(x) ! 5, lim# f(x) ! 5, and f (2) ! 0, find
xS2
xS2
lim f(x), if it exists. Explain your conclusions.
xS2
(b) If lim$ f(x) ! 3, lim# f(x) ! 0, and f (0) ! 0, find
xS0
xS0
lim f(x), if it exists. Explain your conclusions.
xS0
xS10
53. Advertising and sales Suppose that the daily sales S
(in dollars) t days after the end of an advertising campaign are
S ! S(t) ! 400 $
(b) Find lim S(t).
xS100
tS7
54. Advertising and sales Sales y (in thousands of dollars) are related to advertising expenses x (in thousands
of dollars) according to
(b) Find lim$ y(x).
(a) Find lim y(x).
xS10
xS0
55. Productivity During an 8-hour shift, the rate of
change of productivity (in units per hour) of children’s
phonographs assembled after t hours on the job is
r(t) !
128(t2 $ 6t)
,
(t2 $ 6t $ 18)2
56. Revenue If the revenue for a product is R(x) !
100x # 0.1x2, and the average revenue per unit is
R(x) !
xS100
R(x)
x
and
C(p) !
xS40
x
4
$ 30 $ ,
x
4
4 ) x ) 100
R(x)
, x'0
x
(b) lim$
xS0
R(x)
.
x
57. Cost-benefit Suppose that the cost C of obtaining
water that contains p percent impurities is given by
find lim P(x).
S(x) !
0)t)8
(a) Find lim r(t).
(b) Find lim# r(t).
tS4
tS8
(c) Is the rate of productivity higher near the lunch
break (at t ! 4) or near quitting time (at t ! 8)?
P(x) ! 92x # x2 # 1760
51. Sales and training The average monthly sales volume
(in thousands of dollars) for a firm depends on the number of hours x of training of its sales staff, according to
200x
, x*0
x $ 10
y ! y(x) !
find (a) lim
50. Profit If the profit function for a product is given by
2400
t$1
tS14
49. Revenue The total revenue for a product is given by
where x is the number of units sold. What is lim R(x)?
x*4
(b) Find lim S(x).
xS4
A P P L I C AT I O N S
R(x) ! 1600x # x2
9
x
$ 10 $ ,
x
4
(a) Find lim$ S(x).
(a) Find S(0).
(c) Find lim S(t).
lim (1 $ a)1!a
xS100#
52. Sales and training During the first 4 months of
employment, the monthly sales S (in thousands of dollars) for a new salesperson depend on the number of
hours x of training, as follows:
45. Use values 0.1, 0.01, 0.001, 0.0001, and 0.00001 with
your calculator to approximate
aS0
lim S(x).
(b) Find
xS4
120,000
# 1200
p
lim C(p), if it exists. Interpret this result.
pS100#
(b) Find lim$ C(p), if it exists.
pS0
(c) Is complete purity possible? Explain.
(a) Find
9.1 Limits ●
58. Cost-benefit Suppose that the cost C of removing
p percent of the particulate pollution from the smokestacks of an industrial plant is given by
C(p) !
730,000
# 7300
100 # p
(a) Find lim C(p).
pS80
(b) Find lim # C(p), if it exists.
pS100
(c) Can 100% of the particulate pollution be removed?
Explain.
59. Federal income tax Use the following tax rate schedule for single taxpayers, and create a table of values that
could be used to find the following limits, if they exist.
Let x represent the amount of taxable income, and let
T(x) represent the tax due.
lim T(x)
(a)
xS29,050#
lim T(x)
(b)
xS29,050$
(c) lim T(x)
xS29,050
Schedule X––Use if your filing status is Single
If your taxable
income is:
The tax is:
of the
amount
over ––
But not
over ––
Over ––
$0
7,150
$7,150
29,050
10%
$715.00 + 15%
$0
7,150
29,050
70,350
70,350
146,750
4,000.00 + 25%
14,325.00 + 28%
29,050
70,350
146,750
319,100
319,100
35,717.00 + 33%
92,592.50 + 35%
146,750
319,100
61. Municipal water rates The Corner Water Corp. of
Shippenville, Pennsylvania has the following rates per
1000 gallons of water used.
Usage (x)
Cost per 1000 Gallons
(C(x))
First 10,000 gallons
Next 110,000 gallons
Over 120,000 gallons
$7.98
6.78
5.43
If Corner Water has a monthly service fee of $3.59,
write a function C ! C(x) that models the charges
(where x is thousands of gallons) and find lim C(x)
xS10
(that is, as usage approaches 10,000 gallons).
62. Phone card charges A certain calling card costs
3.7 cents per minute to make a call. However, when the
card is used at a pay phone there is a 10-minute charge
for the first minute of the call, and then the regular
charge thereafter. If C ! C(t) is the charge from a pay
phone for a call lasting t minutes, create a table of
charges for calls lasting close to 1 minute and use it to
find the following limits, if they exist.
(a) lim# C(t)
(b) lim$ C(t)
(c) lim C(t)
tS1
tS1
tS1
Dow Jones Industrial Average The graph in the following figure shows the Dow Jones Industrial Average
(DJIA) at 1-minute intervals for Monday, October 5,
2004. Use the graph for Problems 63 and 64, with t as
the time of day and D(t) as the DJIA at time t.
lim
D(t), if it exists. Explain what this
63. Estimate
tS9:30AM$
limit corresponds to.
lim D(t), if it exists. Explain what this
64. Estimate
tS4:00PM#
limit corresponds to.
Source: Internal Revenue Service, 2004, Form 1040 Instructions
60. Parking costs The Ace Parking Garage charges $5.00
for parking for 2 hours or less, and $1.50 for each extra
hour or part of an hour after the 2-hour minimum. The
parking charges for the first 5 hours could be written as
a function of the time as follows:
$5.00
$6.50
f(t) ! d
$8.00
$9.50
(a) Find lim f(t), if it exists.
tS1
(b) Find lim f(t), if it exists.
tS2
595
if 0 & t ) 2
if 2 & t ) 3
if 3 & t ) 4
if 4 & t ) 5
Source: Bloomberg Financial Markets, The New York
Times, October 6, 2004. Copyright © 2004. The
New York Times Co. Reprinted by permission.