7.1 Investigating Limits—Methods of Inquiry and a Definition 251beginning of this section, we will not be applying this definition much in practice; it is the meaningbehind the definitio
Trang 17.1 Investigating Limits—Methods of Inquiry and a Definition 251
beginning of this section, we will not be applying this definition much in practice; it is the
meaningbehind the definition that is of primary importance to us
f
x
L +
L –
L
3 – δ 3 + δ δ
δ 3
f
x L
3 – δ 3 3 + δ
For all x within of 3,
f (x) is within of L
For all x within of 3,
f (x) is well within of L
∈
L + ∈
∈
L – ∈
∈
∈
Figure 7.7
EXAMPLE 7.4 Consider the function f (x) =
x
2, for x = 3, undefined, for x = 3
3
f
x
f (x)
Figure 7.8
We can write this function compactly as f (x) =x(x−3)2(x−3), becausex(x−3)2(x−3)=x2 for x = 3 and
is undefined at x = 3 Find limx→3 x(x−3)2(x−3)
SOLUTION
lim x→3
x(x − 3) 2(x − 3) = limx→3
x
2 (provided x = 3)
=3 2 The argument given in Example 7.3 holds without alteration since we always worked with the condition x = 3 A single hole in the graph makes no difference in the limit In fact, inserting any finite number of holes in the graph of a function has no effect on the computation of limits
EXAMPLE 7.5 Let f (x) =
x
2, for x = 3,
2, for x = 3
Find limx→3f (x)
Trang 2f
x
f (x)
Figure 7.9
SOLUTION Again limx→3f (x) = 32 since f (x) =x2 for x = 3 and the arguments given above, both
formal and informal, hold here as well
Notice that we’ve established that limx→3f (x) = 32 for each of the functions f (x)
drawn below The limit tells us about the behavior of f near x = 3 but not at x = 3.
3
f
x
3
f
x
3
f
x
Figure 7.10
EXAMPLE 7.6 Find limx→0 4x+xx 2
We’re interested in the behavior of f (x) = 4x+xx 2 =(4+x)xx for x near zero but not at
x = 0 For x = 0,(4+x)xx = 4 + x, so limx→0 (4+x)x
x = limx→04 + x = 4
More formally, we must show that for any positive , |f (x) − 4| < provided |x| is small enough (and nonzero) For x = 0, we have:
|f (x) − 4| =
(4 + x)x
= |4 + x − 4| = |x|,
so if 0 < |x| < , then |f (x) − 4| < as well
4
f
x
f (x) = (4 + x) x = 4 + x for x ≠ 0
undefined for x = 0 x
Figure 7.11
The limit in Example 7.6 is the limit that we would compute if we were computing the derivative of g(x) = x2at x = 2
Trang 37.1 Investigating Limits—Methods of Inquiry and a Definition 253
g(2) = lim h→0
g(2 + h) − g(2) h
= lim h→0
(2 + h)2− 4 h
= lim h→0
4 + 4h + h2− 4 h
= lim h→0
(4 + h)h
h This is the limit given in Example 7.6
= lim h→04 + h
= 4
(2, 4)
(2 + h, (2 + h)2 )
g (x) = x2
g
x
Figure 7.12
EXERCISE 7.1 Show that
lim x→3
2x2− 6x
x − 3 = 6 using the formal definition of a limit
Try this first on your own The answer is provided below.
Answer
Let f (x) =2x2−6x
x−3 We must show that for any positive , |f (x) − 6| < provided |x − 3|
is small enough but nonzero If |x − 3| = 0 then
|f (x) − 6| =
2x(x − 3)
x − 3 − 6
= |2x − 6| = 2|x − 3|
Therefore, |f (x) − 6| < provided 0 < |x − 3| <2
EXAMPLE 7.7 Interpret limx→3 x2−9
x−3 as the derivative of some function K(x) evaluated at x = a (a) Determine K(x) and the value of a
(b) Evaluate the limit
Trang 4SOLUTION (a) A derivative is the limit of the slope of the secant lines, so we want to interpretxx−32−9as
K
x, the slope of the line through the points (x, x2)and (3, 9)
We see that the limit is being taken as x approaches 3, so the point (x, x2) is approaching (3, 9) along the curve K(x) = x2
(3, 9)
(x, x2 )
K
K (x) = x2
x
Figure 7.13
K(x) = x2and a = 3 That is, the derivative of x2at x = 3 is limx→3 x
2
−9 x−3 (b) There are several ways of evaluating this limit
Method (i): In part (a) we showed that limx→3 x2−9
x−3 is the derivative of x2at x = 3
In Chapter 6 we showed that the derivative of x2is 2x Evaluating this at x = 3 gives limx→3xx−32−9= 6
Method (ii): Following the reasoning of the previous few examples, we can observe that
x2− 9
x − 3 =
(x − 3)(x + 3)
x + 3, for x = 3, undefined, for x = 3
So
lim x→3
x2− 9
x − 3 = limx→3
(x − 3)(x + 3)
x − 3
= lim x→3x + 3 since x − 3
x − 3= 1 for x = 3
= 6
Alternative ways of approaching the limit:Suppose you’re staring at this limit and feel
at a loss, or for some reason are feeling queasy about your answer Or maybe you just want to double-check your work You can investigate the limit graphically or investigate it numerically
6
3
3
x2 – 9
x– 3
y
y =
x
x
= x + 3 for x ≠ 3
undefined for x = 3
2.99 2.999 3.001 3.01
5.99 5.999 6.001 6.01
Figure 7.14
Trang 57.1 Investigating Limits—Methods of Inquiry and a Definition 255
These methods lead you to conjecture that the limit is 6
EXAMPLE 7.8 Find limx→∞x2
SOLUTION We are being asked about the behavior of x2 as x grows without bound As x grows
without bound, x2grows without bound; about these facts there should be no controversy The problem is how we should answer the question Reasonable people may differ Some mathematicians don’t like to write
lim x→∞x2= ∞
After all, infinity is not something that f (x) = x2 can snuggle up to; f (x) can’t get arbitrarily close to infinity On the other hand, “as x → ∞, x2→ ∞” is a reasonable shorthand for “as x grows without bound, x2grows without bound.” Therefore in this text
we use the convention that
lim x→∞x2= ∞
is a shorthand for “as x grows without bound, x2 grows without bound.” Similarly, limx→af (x) = ∞ is shorthand for “as x approaches a, f (x) grows without bound”
Before moving on from the definition of a limit, let’s take a moment to debunk a common misconception with the next example
EXAMPLE 7.9 Find limx→0 xx
The graph ofxx is given below xx= 1 for x = 0 and is undefined for x = 0
lim x→0
x
x = lim x→01
= 1
1
x x
y
y =
x
= 1 for x ≠ 0
undefined for x = 0
Figure 7.15
From this example we see that limx→af (x) = L does not mean “as x gets closer and closer to a, f (x) gets closer and closer to L but never reaches L”; in Example 7.9 f (x) doesreach L = 1 as x approaches 0 It is possible (but not required) that f will hit the value
Lany number of times
EXERCISE 7.2 Show, using the methods of Example 7.3, that limx→3k = k for any constant k Conclude
that limx→ck = k for c and k any constants
EXERCISE 7.3 Show, using the methods of Example 7.3, that limx→cx = c, where c is any constant
Trang 6P R O B L E M S F O R S E C T I O N 7 1
1 Evaluate the following limits; then discuss limx→∞bxfor b > 0
(a) lim x→∞(1.1)x (b) lim
x→∞(0.9)x (c) lim
x→0(1.1)x (d) lim
x→−∞(1.1)x (e) lim
x→−∞(0.9)x
2 We can define −12
n for any positive integer n, but not for every real number For instance, −211/2=
−12, which is not defined in real numbers We’ll write lim
n→∞
−12
n
= L if
−12
n can be made arbitrarily close to L for all positive integers
nsufficiently large
(a) Find lim
n→∞
−1 2
n , where n takes on only positive integer values
(b) Which two of the following statements are true? Explain
i lim n→∞(−2)n
= ∞
ii lim n→∞(−2)n= −∞
iii lim n→∞(−2)ndoes not exist
iv lim
v lim
3 Each of the following limits are of the form limx→a f (x) Evaluate the limit and sketch
a graph of f on some interval including a Make it clear from your sketch whether or not f is defined at a
(a) lim h→−2
(h − 3)(h + 2)
x2− 25
x + 5 (c) lim
x→−5
x2− 25
t2+ πt t (e) lim
h→0
hk + h2
h , where k is a constant (f ) lim
w→2
(w − 3)(w + 1)(w − 2) 3w − 6
4 Discussion Question Consider limx→∞(1 + 1x)x This is a very important limit
(a) What is your guess for this limit? You are not expected to guess the right answer Once you complete part (b) you will see that the problem is subtle.
(b) Use a calculator or computer to investigate the limit, graphically and numerically
Give a revised estimate of this limit We will return to this limit in Chapter 15.
In Problems 5 through 13, graph f and evaluate the limit(s).
5 f (x) = x2 + 3; lim
x→2f (x)
6 f (x) = πx − 4;
(a) lim x→0f (x) (b) lim
x→1f (x) (c) lim
x→∞f (x)
Trang 77.1 Investigating Limits—Methods of Inquiry and a Definition 257
7 f (x) = |x − 2|;
(a) lim
x→0f (x) (b) lim
x→2f (x)
8 f (x) =x
2− 2x
x − 2 ;
(a) lim
x→0f (x) (b) lim
x→2f (x)
9 f (x) = 5x + 1, x = 2
7, x = 2;
(a) lim
x→1f (x) (b) lim
x→2f (x)
10 f (x) = (x + 2)(x
2
− x) x(x − 1) ;
(a) lim
x→0f (x) (b) lim
x→1f (x)
11 f (x) = x
2− 3x − 4
x + 1 ;
(a) lim
x→1f (x) (b) lim
x→−1f (x)
12 f (x) = |x|, x = 3
0, x = 3;
(a) lim
x→0f (x) (b) lim
x→3f (x) (c) lim
x→−∞f (x)
13 f (x) = x
2− 9
x + 3;
(a) lim
x→3f (x) (b) lim
x→−3f (x)
In Problems 14 through 18, evaluate the limit by interpreting it as the derivative of some function f evaluated at x = a Specify f and a, then calculate the limit.
14 lim
x→2
ex− e2
x − 2
15 lim
h→0
√
9 + h − 3
h
16 lim
h→0
e1+h− e
h
17 lim
h→0
√
7 + h −√7
h
18 lim
h→0
eh− 1
h
In Problems 19 and 20, give an example of a function having the set of characteristics specified.
Trang 819 (a) lim x→5f (x) = 7; f (5) = 7 (b) lim
x→5g(x) = 7; g(5) = 8
20 (a) lim x→∞f (x) = ∞; lim
x→−∞f (x) = −∞; lim
x→0f (x) = 1 (b) lim
x→∞g(x) = ∞; lim
x→−∞g(x) = ∞; lim
x→0g(x) = −1
21 Look back at Example 7.6 When approximating the slope of x2at x = 2, we end up with the expression (4h + h2)/ h If we assume h = 0, then we can cancel the h’s, arriving at 4 + h In this problem, we will investigate a function like (4h + h2)/ h Consider the following pay scale for employees at the nepotistic Nelson Nattle Company Let D be the date of hire (we use the start-up date of the Nattle Company
as our benchmark time, D = 0, D measured in years) and let S be the starting annual salary
S(D) =
(15000 + 200D)D(D − 1)
D(D − 1)
50, 0000 for D = 0
35, 000 for D = 1 (a) Sketch the graph of S(D) The graph looks a little weird until you find out that Nelson Nattle is the only person hired at D = 0 and that his son, Nelson Nattle, Jr., is due back from college exactly one year after the Nattle Company’s
start-up date The expression(15000+200D)D(D−1)
D(D−1) is equal to 15000 + 200D as long as
D = 0 and D = 1 For D = 0 and D = 1,(15000+200D)D(D−1)D(D−1) is undefined (b) Nelson has just received a letter from his brother Nathaniel asking for a position
in the company Nathaniel’s projected date of hire is D = 1.5 Nelson is thinking
of offering his brother a starting salary of $40,000 Adjust S(D) to define S(1.5) appropriately
22 Let f (t) =t (3 + t)t (a) Sketch the graph of f (t)
(b) What are the domain and range of f (t)?
7.2 LEFT- AND RIGHT-HANDED LIMITS;
SOMETIMES THE APPROACH IS CRITICAL
EXAMPLE 7.10 How does f (x) =1
x behave as x approaches zero?
SOLUTION Let’s begin by investigating this graphically and numerically
Trang 97.2 Left- and Right-Handed Limits; Sometimes the Approach Is Critical 259
f
f (x) =
x
x
.01 001 0001 –.0001 –.001 –.01
100 1000 10000 –10000 –1000 –100
Figure 7.16
As x approaches zero from the right (i.e., through numbers greater than zero),x1grows without bound We can express this by writing
lim x→0 +
1
x = ∞, where “x → 0+” indicates that only numbers greater than zero are being considered We read “limx→0+” as “the limit as x approaches 0 from the right.”
As x approaches zero from the left (i.e., through numbers less than zero),1x decreases without bound We can express this by writing
lim x→0 −
1
x = −∞, where “x → 0−” indicates that only numbers less than zero are being considered
We can define finite one-sided limits in general as follows.
D e f i n i t i o n s
If the values of f (x) stay arbitrarily close to L provided x is sufficiently close to a but greater than a, we write
lim x→a +f (x) = L and read this as “the limit of f (x) as x approaches a from the right is L.”
Analogously, if the values of f (x) can be made to stay arbitrarily close to L provided x is sufficiently close to a but less than a, then we write
lim x→a −f (x) = L
These definitions can be made more precise in exactly the same manner the definition of limx→af (x) = L was made more precise From that latter definition we see that in order
to have limx→af (x) = L it is necessary that both the left- and right-hand limits be equal
to L as well
Trang 10If lim x→a −f (x) = lim
x→a +f (x), then lim
x→af (x)does not exist
(We will make a stronger statement about the relationship between one-sided and two-sided limits later in this section.) It follows that limx→0x1does not exist
EXAMPLE 7.11 Evaluate limx→0+ |x|
x, limx→0− |x|
x , and limx→0|x|x
SOLUTION The basic strategy for dealing with absolute values is to split the problem into two cases, one
where the quantity inside the absolute value is negative and the other where this quantity is nonnegative
If x > 0, |x|
x =xx = 1;
if x < 0, |x|
x =−xx = −1;
if x = 0, |x|x is undefined
1
–1
|x|
x
y
y =
x
Figure 7.17
limx→0+ |x|
x = limx→0 +1 = 1, limx→0− |x|
x = limx→0 −−1 = −1, limx→0 |x|x does not exist because the left- and right-hand limits are unequal
EXAMPLE 7.12 Evaluate limx→0+|x|, limx→0 −|x|, and limx→0|x|
SOLUTION
If x ≥ 0, |x| = x;
if x < 0, |x| = −x
y
y = |x |
x
Figure 7.18