This is not the only situation in which a limit does not exist.. limx→0gxdoes not exist since in any open interval around zero, no matter how small, there will be both rational and irrat
Trang 1limx→0+|x| = limx→0 +x = 0, limx→0−|x| = limx→0 −(−x) = 0, limx→0|x| = 0
EXAMPLE 7.13 Let f (x) = |x|.
(a) Sketch the graph of f and sketch the graph of f (b) Evaluate f(1) and f(−2)
(c) Verify that f(0) is undefined
SOLUTION
(a) i graph of f (x) = |x| ii graph of f(x) =dxd |x|
1
–1
f ′ f
x x
f (x) = |x|
Figure 7.19
(b) For x > 0 f (x) = x, so f(1) = 1
For x < 0 f (x) = −x, so f(−2) = −1
(c) Approaching the problem from a graphical viewpoint, we notice that the graph of f does not appear locally linear at x = 0 No matter how much the region around x = 0
is magnified the graph has a sharp corner there, so it will never look like a straight line The slope of f is −1 if x approaches zero from the left and +1 if x approaches zero from the right
Analytically we can argue similarly
f(0) = lim h→0
f (0 + h) − f (0) h
= lim h→0
|h| − |0|
h
= lim h→0
|h|
h
This is the limit we looked at in Example 7.11 We concluded that limh→0|h|h does not exist Therefore dxd|x|
x=0, the derivative of |x| evaluated at x = 0, does not exist
EXAMPLE 7.14 Find limt →0 1
t 2
SOLUTION limt →0+ 1
2 = +∞; likewise, limt →0 − 1
2 = +∞
Trang 2262 CHAPTER 7 The Theoretical Backbone: Limits and Continuity
Therefore we say limt →0 1
t 2 = +∞
f
t
f (t) = 1
t2
Figure 7.20 The relationship between left- and right-hand limits and two-sided limits can be stated most succinctly as follows
lim x→af (x) = L if and only if lim
x→a +f (x) = lim
x→a −f (x) = L
The limit as x approaches a of f (x) is L if and only if the left- and right-hand limits are both equal to L
In this section we’ve seen that if limx→a−f (x) = L1 and limx→a+f (x) = L2 but
L1= L2, then limx→af (x)does not exist This is not the only situation in which a limit does not exist Consider, for example, the function g(x) that is 1 when x is rational and −1 when x is irrational limx→0g(x)does not exist since in any open interval around zero, no
matter how small, there will be both rational and irrational numbers In this case neither the left- nor right-hand limit exists Similarly, limx→∞g(x)does not exist, because for any
N, no matter how big N is, there will always be both rational and irrational numbers larger than it
P R O B L E M S F O R S E C T I O N 7 2
1 Let f be the first-class postage function for 1997 given in Example 2.3 The input of the function is W , where W is the weight, in ounces, of the item to be mailed and the output is the price of the postage Return to Chapter 2 for details Find the following
(a) lim
W →1 − f (W ) (b) lim
W →1 + f (W ) (c) lim
W →1 f (W ) (d) lim
W →1.5 f (W ) (e) lim
W →12 + f (W ) (g) lim
W →12 − f (W ) (h) lim
W →12 f (W )
In Problems 2 through 7, evaluate the limits A graph may be useful.
2 (a) lim x→2 −
1
x→2 +
1
x→2
1 x−2
3 (a) lim x→1 −
1 (x−1) 2 (b) lim
x→1 +
1 (x−1) 2 (c) lim
x→1
1 (x−1) 2 (d) lim
x→−1
1 (x−1) 2
4 f (x) = |x−3|x−3
Trang 3(a) lim
x→0f (x) (b) lim
x→4f (x) (c) lim
x→3 +f (x) (d) lim
x→3 −f (x) (e) lim
x→3f (x)
5 f (x) = 3x + 4, x < 0
2x + 4, x ≥ 0
(a) lim
x→1f (x) (b) lim
x→−2f (x) (c) lim
x→0 +f (x) (d) lim
x→0 −f (x) (e) lim
x→0f (x)
6 f (x) = πx + 1, x > 0
π x − 1, x ≤ 0
(a) lim
x→1π
f (x) (b) lim
x→−π1
f (x) (c) lim
x→0f (x)
7 f (x) = x
2+ 3, x ≥ 1
2x + 2, x < 1
(a) lim
x→0f (x) (b) lim
x→1f (x) (c) lim
x→2f (x)
8 Let f be the function defined by f (x) = x + 1, for x not an integer,
0, for x an integer
(a) Sketch f
(b) Find the following limits
i lim
x→1.5f (x) ii lim
x→2 f (x) iii lim
x→0 f (x) (c) For what values of c is limx→c f (x) = c + 1? Have you excluded any values of c? If so, which ones and why? Explain
9 Let g(x) be the function defined by g(x) =
x2, for x > 2,
x, for x ≤ 2
(a) Find the following
i lim
x→2 + g(x) ii lim
x→2 − g(x) iii lim
x→2 g(x) (b) h(x) is defined by h(x) =
x2, for x > 2,
mx, for x ≤ 2
Find the value of m that will make f (x) a continuous function (A graph is highly recommended.) Using the value of m you’ve found, determine limx→2 h(x)
10 Let f be the function given by
f (x) =
1 x+7, for x ≤ −5,
x + 4, for −5 < x < −2,
|x|, for −2 ≤ x ≤ 2,
2−x+ 1, for x > 2
Trang 4264 CHAPTER 7 The Theoretical Backbone: Limits and Continuity
A sketch of f is provided below
x f
1
1 –1 –1 –2 –3 –4 –5 –6
2
3 4 5 6 7 8 9
Evaluate the limits below
(a) limx→−∞ f (x) (b) limx→−7 f (x) (c) limx→−5+ f (x) (d) limx→−5 f (x) (e) limx→−2 f (x) (f ) limx→2+ f (x) (g) limx→∞ f (x) (h) limx→0 f (x)
11 Let f (x) be the function defined and graphed in Problem 10
(a) Sketch a graph of f(x)
(b) Where is fundefined?
(c) Evaluate the following
i limx→∞f(x) ii limx→−2−f(x) iii limx→−2+f(x)
iv limx→−2f(x) v limx→−7−f(x) vi limx→−7+f(x)
12 (a) Sketch the graph of f (x) = 2|x − 2|
(b) What is f(0)? What is f(4)? What is f(2)?
(c) Find limh→0+f (2+h)−f (2)
h and limh→0− f (2+h)−f (2)
Do your answers make sense to you?
(d) On a separate set of axes graph f(x)
Each of the statements in Problems 13 through 16 is false Produce a counterexample
to show that the statement is false.
13 If f (3) = 7, then limx→3f (x) = 7
14 If f (2) is undefined, then limx→2f (x)is undefined
15 If limx→0+f (x)and limx→0−f (x)both exist, then limx→0f (x)exists
16 If limx→5f (x) = 2 then f (5) = 2
For Problems 17 through 21, use the graph given to determine each of the following.
(a) limx→1+f (x) (b) limx→1−f (x) (c) limx→1f (x)
Trang 517 18 19.
f
x
1
f
x
1 2
f
x
1 –1 –2
f
x
1
f
x
7.3 A STREETWISE APPROACH TO LIMITS
Suppose you and your trusty calculator are strolling down the street one night As you pass
a dark alley the Math Mugger jumps you, throwing you a nasty limit A quick answer is demanded Although normally cool and level-headed, you start to panic You barely heard the question, but the words “zero,” “one,” “infinity,” and “does not exist” are at the tip of your tongue You try to calm yourself, asking boldly, “Could you repeat the question, please?” Let’s suppose the question thrown at you is “What is limx→1xx−13−1?” First let’s look
at potholes to avoid Then we’ll supply some calculus street-survival tactics Finally, we’ll determine whether using these survival tactics would, in this particular instance, get you out of the bind you’re in
Potholes and Mudholes with Crocodiles
In your panic you might look at limx→1 x3−1
x−1 and think, “when x → 1 the numerator tends toward 0, and if the numerator is 0 then the fraction is 0, so the answer is 0.” Alternatively, you might think, “when x → 1 the denominator tends toward 0, and when the denominator goes to 0 the limit does not exist or is infinite.” If you’ve had these two thoughts in rapid succession, you might put them together and think, “so we have 00, which is undefined whenever it’s not 1.”3
There are serious problems with these lines of thought While it is true that both the numerator and the denominator of this fraction are approaching zero, we’re interested in
3
Trang 6266 CHAPTER 7 The Theoretical Backbone: Limits and Continuity
what this ratio looks like as x → 1 Think about this carefully; every derivative we calculate
looks like lim x→0 f x and therefore looks like 00, yet this limit could work out to be any
number, depending upon the function To assert that every limit of the form00is either 1 or
does not exist is equivalent to asserting that all derivatives are 1 or undefined, an assertion
you know is patently false!
Another approach you might take to limx→1x3−1
x−1 is to simplify the fractionx3−1
x−1 That’s
a solid idea If you can factor x3− 1, you’re in great shape and don’t need help It’s entirely possible, however, that you know how to factor the difference of two perfect squares but not
of two perfect cubes Do not resort to wishful-thinking algebra If you’re not sure of your algebra, you’ve got means at your disposal to check it
If you’re not sure of a factorization, multiply out to determine whether it works
If you think you can simplify without factoring (you can’t), do a spot-check on yourself For example, whatever you simplify this to, when you evaluate at x = 2 you should get
23−1 2−1 = 7
Suppose you’re stuck; you can’t figure out how to simplify the fraction x3−1
x−1 and your panic level is mounting.4What you need are street-survival skills
Street-Survival Tactics
The survival tactics are simple and can be applied in a large variety of situations, not just limit calculations in back alleys
Investigate the situation graphically
Investigate the situation numerically
The potholes identified above are analytic and algebraic missteps Don’t put on analytic blinders; view the problem from different angles
Graphical and Numerical Investigations
We’re interested in limx→1x3−1
x−1, so we want the graph of f (x) = x3−1
x−1 around x = 1 For starters we might graph on the interval [0, 2] and then zoom in around x = 1 To make sure
we have the graph in the viewing window, we can begin by letting the range of y-values be from f (0) =00−13−1= 1 to f (2) =22−13−1 = 7 We can zoom in around x = 1 as many times
as we like (As pointed out in Chapter 2, although f (x) is undefined at x = 1, the graph
on the calculator may not reflect this.) Use the “trace” key to trace along the curve near
x = 1
4
Trang 7x = 997 x = 1.002
y = 3.008
y = 2.990 tracer
Figure 7.21
It is reasonable to conjecture that as x → 1,x3−1
x−1 → 3
Another approach is to use your calculator to calculate values of x3−1
x−1 for values of
x approaching 1 (For instance, you could try 0.99, 0.9997, 1.0001, 1.00003, etc., or you could get your calculator to produce a table of values.) Again, your investigation would lead you to conjecture that as x → 1, x3−1
x−1 → 3
So if the fellow in the back alley demands an answer immediately, tell him your best guess is 3
Is this answer correct? Actually it is x3− 1 can be factored into (x − 1)(x2+ x + 1), allowing us to do the following.5
lim x→1
x3− 1
x − 1 = limx→1
(x − 1)(x2+ x + 1)
x − 1
= lim x→1(x2+ x + 1)
= 3 (Alternatively, this limit can be interpreted as the derivative of x3at x = 1 and calculated
as limh→0 (1+h)h3−13 Do this calculation as an exercise.)
Cautionary Notes
If you’re going to use street-survival tactics, you’re going to have to learn how to watch
your back The tactics recommended will often get you near the actual answer (not always,
but frequently, if used wisely), but there is no reason to believe that they will lead you to the exact answer For instance, in the last example at some point we simply guessed the answer was 3 (as opposed to some number very, very close to 3) If the actual answer was irrational, the calculator would probably not tell us this
There are other limitations inherent in your trusty machine; sometimes the calculator can lead you astray Suppose, for instance, we are interested in the derivative of 2xat x = 0
It is given by
lim h→0
2h
− 1
h
5
Trang 8268 CHAPTER 7 The Theoretical Backbone: Limits and Continuity
f
x
(0, 1)
f (x) = 2 x (h, 2 h)
tangent line at x = 0
Figure 7.22
We can approximate this derivative by2hh−1 for h small The approximation should be better and better the smaller h is If h is positive, the value obtained should be greater than the actual limit (See Figure 7.20.) The following are data collected from a calculator
− 1 h
10−2= 0.01 0.6955550057
10−3= 0.001 0.693387463
10−4= 0.0001 0.6931712
10−5= 0.00001 0.6931496
10−6= 0.000001 0.693147
10−7= 0.0000001 0.69315
10−8= 0.00000001 0.6931
10−9= 0.000000001 0.693
10−10 0.69
10−11 0.7
10−12 1
10−13 0
10−14 0
Is the limit actually 0, not ≈ 0.693? The actual value of limh→02
h −1
h is an irrational number This irrational number is approximately 0.6931471806 What is going on here?
The information supplied by the calculator looks reasonable up until h = 10−6 For the values of h smaller than this, the difference between 2hand 1 is so tiny that the calculator
is essentially losing one digit of information for each subsequent entry up through 10−12
By h = 10−13and h = 10−14, the difference 2h
− 1 is so miniscule that the calculator just rounds the difference off to 0
This phenomenon is not a quirk particular to this calculator, nor is it particular to the function 2x Similar issues arise whenever calculating a limit that can be interpreted
as a derivative So what should you do? A streetwise person is not an extremist When investigating limx→af (x), try values of x close to a but not too close And then make your guess and take your chances You’ll win some and you’ll lose some But frequently, thoughtful graphical and numerical investigation will put you in the vicinity of the actual answer
Trang 9Question: When is limx→af (x)simply equal to f (a)?
In the language of the streets, when can you just plug in x = a and get the right answer even though taking the limit means x can’t really be equal to a?6
This is a great question Think about it for a minute We’ll recap Examples 7.3, 7.4, and 7.5 to give you food for thought
f
x
3
f
x
3
2
f (x) = x2
lim f(x) = 32
x→3
f (x) =
f (x) =
x (x–3)
2 (x–3) lim f(x) = 32
x
2
lim f(x) = 3
2
f
x
3
for x ≠ 3 for x = –3
Figure 7.23
If f is continuous at x = a then this ‘method’ will work This brings us back from the streets to define continuity using limits
P R O B L E M S F O R S E C T I O N 7 3
1 Is limx→∞(1 +1x)xfinite? If so, find two consecutive integers, one smaller than this limit and the other larger We will return to this limit later in the course
In Problems 2 and 3, find the limit.
2 lim
x→3
2x3− 8x2+ 5x + 3
x − 3
3 lim
x→0
√
4 + x − 2
x
4 (a) Suppose you are interested in finding limx→2f (x), where the function f (x) is explicitly given by a formula What approaches might you take to investigate this limit?
(b) Suppose you’re now interested in finding limx→∞f (x), where f (x) again is explicitly given by a formula What approaches might you take to investigate this limit?
6
Trang 10270 CHAPTER 7 The Theoretical Backbone: Limits and Continuity
7.4 CONTINUITY AND THE INTERMEDIATE
AND EXTREME VALUE THEOREMS
Our intuitive notion is that a function is continuous if its graph can be drawn without lifting pencil from paper For f to be continuous at x = a the values of f (x) must be close to f (a) for x near a To define continuity we’ll begin by looking at the type of behavior we are trying to rule out The function sketched below is discontinuous at x = −3, x = −1, x = 0,
x = 1, and x = 5.2
f
x
Figure 7.24
To begin with, we might insist that if f is to be continuous at x = a, then
lim x→a −f (x) = lim
x→a +f (x),
where this limit is finite For the function shown in Figure 2.24, this rules out the jump discontinuities at x = −3, x = 1, and x = 5 but doesn’t take care of the removable
discon-tinuities at x = −1 and x = 0 The latter reminds us that f (a) must be defined and equal to the left- and right-hand limits In other words, the following conditions must be satisfied:
i f must be defined at x = a,
ii left- and right-hand limits at x = a must be equal, iii f (a) must be equal to limx→af (x)
We can put this more succinctly
D e f i n i t i o n
The function f is continuous at x = a if
lim x→a −f (x) = lim
x→a +f (x) = f (a)
Equivalently, f is continuous at x = a if
lim x→af (x) = f (a)
f is continuous on an open interval if f is continuous at every point in the interval.7 There are two theorems concerning continuous functions that we will use repeatedly
7
An open interval can be of the form (a, b), (a, ∞), (−∞, b) or (−∞, ∞).