2.5 SECTION CONTINUITY
||||
119
2.5
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 denition 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.)
1
DEFINITION A function f is continuous at a number a if
lim f x
x la
fa
N
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. y
Notice that Denition l implicitly requires three things if f is continuous at a:
1. f a is dened (that is, a is in the domain of f ) 2. lim f x exists
x la x la
y=
f (a)
3. lim f x
fa
approaches f(a).
0
As x approaches a, FIGURE 1
a
x
The denition 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 sufciently small. If f is dened near a (in other words, f is dened 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 persons 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-
y
ous? Why?
S O L U T I O N It looks as if there is a discontinuity when a
0
1
2
3
4
5
x
FIGURE 2
1 because the graph has a break there. The ofcial reason that f is discontinuous at 1 is that f 1 is not dened. The graph also has a break when a 3, but the reason for the discontinuity is different. Here, f 3 is dened, 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 dened 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. Now lets see how to detect discontinuities when a function is dened by a formula.
M
120
||||
C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S
V EXAMPLE 2
Where are each of the following functions discontinuous? x x 2 x x 2 2 if x if x 2 2 2 (b) f x 1 x2 1 x if x if x 0 0
(a) f x
x2
x (c) f x 1
SOLUTION
2
(d) f x
(a) Notice that f 2 is not dened, so f is discontinuous at 2. Later well see why f is continuous at all other numbers. (b) Here f 0 1 is dened but
xl0
lim f x
xl0
lim
1 x2
does not exist. (See Example 8 in Section 2.2.) So f is discontinuous at 0. (c) Here f 2 1 is dened and lim f x
x l2
lim
x l2
x2 x
x 2
2
lim
x l2
x
2x x2 f2
1
lim x
x l2
1
3
exists. But lim f x
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.) M Figure 3 shows the graphs of the functions in Example 2. In each case the graph cant 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 redening f at just the single number 2. [The function t x x 1 is continuous.] The discontinuity in part (b) is called an innite discontinuity. The discontinuities in part (d) are called jump discontinuities because the function jumps from one value to another.
y y y y
1 0 1 2 x
1 0 x
1 0 1 2 x
1 0 1 2 3 x
(a) =
-x-2 x-2
1 if x0 (b) = 1 if x=0
(c) =
-x-2 if x2 x-2 1 if x=2
(d) =[ x ]
F I G U R E 3 Graphs of the functions in Example 2
SECTION 2.5 CONTINUITY
||||
121
2
DEFINITION A function f is continuous from the right at a number a if
x la
lim f x
fa
and f is continuous from the left at a if
x la
lim f x
fa
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
x ln
lim f x
x ln
lim x n
n 1
fn fn
M
but
x ln
lim f x
x ln
lim x
3 DEFINITION A function f is continuous on an interval if it is continuous at every number in the interval. (If f is dened 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
1
interval
S O L U T I O N If
1, 1 . 1 a
s1
x 2 is continuous on the
1, then using the Limit Laws, we have
xla
lim f x
xla
lim (1
xla
s1
x2 ) x2 x2
(by Laws 2 and 7) (by 11) (by 2, 7, and 9)
1 1 1 fa
y 1
lim s1 1 a2
s xlim la
s1
Thus, by Denition l, f is continuous at a if
=1- 1-
xl 1
1 and
a
xl1
1. Similar calculations show that lim f x 1 f1
lim f x
1
f
1
-1
0
1
x
so f is continuous from the right at 1 and continuous from the left at 1. Therefore, according to Denition 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
FIGURE 4
1
M
Instead of always using Denitions 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.
122
||||
C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S
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. c f
4
4. f t
5.
f t
if t a
0
P R O O F Each of the ve 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
xla
lim f x
fa
and
xla
lim t x
ta
Therefore
xla
lim f
tx
xla
lim f x
xla
tx
xla
lim f x ta ta
lim t x
(by Law 1)
fa f This shows that f t is continuous at a.
M
It follows from Theorem 4 and Denition 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 dened; 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
xla
lim c0 am
c0 m
(by Law 7)
and
xla
lim x m
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.
SECTION 2.5 CONTINUITY
||||
123
(b) A rational function is a function of the form fx Px Qx
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, M 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 t seconds later 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. x3 2x2 1 . 5 3x x3 2x 2 1 5 3x
5 3
EXAMPLE 5 Find lim
xl 2
S O L U T I O N The function
fx
is rational, so by Theorem 5 it is continuous on its domain, which is {x x Therefore
xl 2
}.
lim
x3
2x2 1 5 3x
xl 2
lim f x 2
3
f 2 22 32
2 1 1 11
M
5
y
P (cos,sin ) 1
0
It turns out that most of the familiar functions are continuous at every number in their domains. For instance, Limit Law 10 (page 101) is exactly the statement that root functions are continuous. 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 denitions 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
(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 (for Section 3.3.
N
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 56 and 57). It follows from part 5 of Theorem 4 that tan x sin x cos x
124
||||
C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S
y
1 _2
3 _
_
2
0
2
3 2
x
F I G U R E 6 y=tan x
is continuous except where cos x 0. This happens when x is an odd integer multiple of 2, so y tan x has innite discontinuities when x 2, 3 2, 5 2, and so on (see Figure 6). The inverse function of any continuous one-to-one function is also continuous. (This fact is proved in Appendix F, but our geometric intuition makes it seem plausible: The graph of f 1 is obtained by reecting the graph of f about the line y x. So if the graph of f has no break in it, neither does the graph of f 1.) Thus the inverse trigonometric functions are continuous. In Section 1.5 we dened the exponential function y a x so as to ll in the holes in the graph of y a x where x is rational. In other words, the very denition of y a x makes it a continuous function on . Therefore its inverse function y log a x is continuous on 0, .
7
N
The inverse trigonometric functions are reviewed in Section 1.6.
THEOREM The following types of functions are continuous at every number in
their domains: polynomials rational functions root functions
trigonometric functions exponential functions
inverse trigonometric functions logarithmic functions
EXAMPLE 6 Where is the function f x
ln x tan 1 x continuous? x2 1
ln x is continuous for x 0 and y tan 1x is continuous on . Thus, by part 1 of Theorem 4, y ln x tan 1x is continuous on 0, . The denominator, y x 2 1, is a polynomial, so it is continuous everywhere. Therefore, by part 5 of Theorem 4, f is continuous at all positive numbers x except where x 2 1 0. So f is continuous on the intervals 0, 1 and 1, . M
EXAMPLE 7 Evaluate lim
xl
S O L U T I O N We know from Theorem 7 that the function y
sin x . 2 cos x
sin x is continuous. The function in the denominator, y 2 cos x, is the sum of two continuous functions and is therefore continuous. Notice that this function is never 0 because cos x 1 for all x and so 2 cos x 0 everywhere. Thus the ratio fx sin x cos x
S O L U T I O N Theorem 7 tells us that y
2
is continuous everywhere. Hence, by denition of a continuous function, lim 2 sin x cos x lim f x f 2 sin cos 0 2 1 0
M
xl
xl
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.
SECTION 2.5 CONTINUITY
||||
125
N
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
In other words,
THEOREM If f is continuous at b and lim t x
x la xla
b, then lim f t x
x la
f b.
lim f t x
f lim t x
xla
(
)
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.
EXAMPLE 8 Evaluate lim arcsin
x l1
1 sx . 1x
S O L U T I O N Because arcsin is a continuous function, we can apply Theorem 8:
lim arcsin
x l1
1 sx 1x
arcsin lim arcsin lim
x l1
x l1
1 sx 1x
(1
1 1 6
1 s sx x ) (1 s x ) sx
arcsin lim
x l1
arcsin
1 2
M
Lets now apply Theorem 8 in the special case where f x tive integer. Then f tx and f lim t x
xla
n st x
n sx , with n being a posi-
(
)
s xlim t x la
n
If we put these expressions into Theorem 8, we get
xla
n lim st x
s xlim t x la
n
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 function f t given by f t x f t x is continuous at a.
This theorem is often expressed informally by saying a continuous function of a continuous function is a continuous function.
P R O O F Since t is continuous at a, we have
xla
lim t x
ta
Since f is continuous at b
t a , we can apply Theorem 8 to obtain
xla
lim f t x
f ta
126
||||
C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S
which is precisely the statement that the function h x is, f t is continuous at a.
V EXAMPLE 9
f t x is continuous at a; that
M
(a) h x
SOLUTION
Where are the following functions continuous? (b) F x sin x 2 ln 1 cos x f t x , where tx x2 and fx sin x
(a) We have h x
2 _10 10
_6
FIGURE 7
y=ln(1+cos 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) We know from Theorem 7 that f x ln x is continuous and t x 1 cos x is continuous (because both y 1 and y cos x are continuous). Therefore, by Theorem 9, F x f t x is continuous wherever it is dened. Now ln 1 cos x is dened when 1 cos x 0. So it is undened when cos x 1, and this happens when x , 3 , . . . . Thus F has discontinuities when x is an odd multiple of and is continuous on the intervals between these values (see Figure 7). M An important property of continuous functions is expressed by the following theorem, whose proof is found in more advanced books on calculus.
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 fa 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 8. Note that the value N can be taken on once [as in part (a)] or more than once [as in part (b)].
y f(b) N f(a) 0 a y f(b) N
y=
y=
c b x
f(a) 0 a c c c b x
y f(a) N f(b) 0 a
FIGURE 8 y=
(a)
(b)
y=N
b
x
FIGURE 9
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 9, then the graph of f cant 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 44). One use of the Intermediate Value Theorem is in locating roots of equations as in the following example.
SECTION 2.5 CONTINUITY
||||
127
V E X A M P L E 10
Show that there is a root of the equation 4x 3 6x 2 3x 2 0
between 1 and 2.
S O L U T I O N Let f x
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 f1 and f2 4 32 6 24 3 6 2 2 1 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
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
M
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 10. Figure 10 shows the graph of f in the viewing rectangle 1, 3 by 3, 3 and you can see that the graph crosses the x-axis between 1 and 2. Figure 11 shows the result of zooming in to the viewing rectangle 1.2, 1.3 by 0.2, 0.2 .
3 0.2
_1
3
1.2
1.3
_3
_ 0.2
FIGURE 10
FIGURE 11
In fact, the Intermediate Value Theorem plays a role in the very way these graphing devices work. A computer calculates a nite 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.
128
||||
C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S
2.5
EXERCISES
is continuous at the number 4. (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
1. Write an equation that expresses the fact that a function f 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 dis-
continuous 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
lim x l 3 2 f x
tx
4, nd t 3 .
5 and
10 12 Use the denition of continuity and the properties of limits to show that the function is continuous at the given number a. 10. f x 11. f x 12. h t
x2 x 2t 1
s7 3t 2 , t3
x, a
4 1
2 x 3 4, a a 1
_4
_2
0
2
4
6
x
13 14 Use the denition of continuity and the properties of limits
to show that the function is continuous on the given interval.
13. f x
2x x 2 s3
3 , 2 x,
2, ,3
4. From the graph of t, state the intervals on which t is
continuous.
14. t x
y
15 20 Explain why the function is discontinuous at the given
number a. Sketch the graph of the function.
15. f x
_4 _2 2 4 6 8 x
ln x 1 x 2 1
2 if x if x 0 0 if x if x if x if x if x 5x 3 3 1 1 0 0 0 if x if x 3 3 1 1
a a
2 1
16. f x
5. Sketch the graph of a function that is continuous everywhere
except at x
3 and is continuous from the left at 3. 4, but is con-
17. f x
e x if x x 2 if x x2 x2 1 x 1
a
0
6. Sketch the graph of a function that has a jump discontinuity
at x 2 and a removable discontinuity at x tinuous elsewhere.
18. f x
a
1
7. A parking lot charges $3 for the rst 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 signicance to someone who parks in the lot.
8. Explain why each function is continuous or discontinuous.
19. f x
cos x 0 1 x2 2x 2
a
0
20. f x
x 6
a
3
(a) The temperature at a specic location as a function of time (b) The temperature at a specic 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
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
x2
x 5x
6
22. G x
3 sx 1
x3
SECTION 2.5 CONTINUITY
||||
129
23. R x 25. L t 27. G t
x2 e
5t
s2 x cos 2 t
4
1
24. h x 26. F x 28. H x
sin x x1 sin
1
41. For what value of the constant c is the function f continuous
on
2
,
? fx c x 2 2 x if x x 3 cx if x 2 2
x
1
ln t
1
cos(e
sx
)
; 29 30 Locate the discontinuities of the function and illustrate by
graphing.
29. y
42. Find the values of a and b that make f continuous everywhere.
1 1 e1 x
30. y
ln tan x
2
fx
x2 4 x2 a x 2 bx 3 2x a b
if x if 2 if x
2 x 3 3
31 34 Use continuity to evaluate the limit.
5 sx 31. lim x l 4 s5 x
33. lim e x
x l1
2
43. Which of the following functions f has a removable disconti-
32. lim sin x
xl
sin x x2 3x 2 4 6x
nuity at a ? If the discontinuity is removable, nd a function t that agrees with f for x a and is continuous at a. (a) f x (b) f x (c) f x x4 x x3 x sin x , 1 , 1 x2 2 a a 2x , 1 a 2
x
34. lim arctan
x l2
35 36 Show that f is continuous on 35. f x
,
.
x 2 if x sx if x sin x if x cos x if x
1 1 4 4
44. Suppose that a function f is continuous on [0, 1] except at
36. f x
37 39 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 .
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 doesnt satisfy the hypothesis).
45. If f x
37. f x
1 2 x
x2 x 2
2
if x if 0 if x
0 x 2 1 x 3 0 x 1
x 2 10 sin x, show that there is a number c such that f c 1000. the equation f x explain why f 3 6 are x 6. 1 and x 4. If f 2 8,
2
46. Suppose f is continuous on 1, 5 and the only solutions of
38. f x
x1 if x 1x if 1 sx 3 if x x ex 2 2 if x if 0 x if x
3
47 50 Use the Intermediate Value Theorem to show that there is
a root of the given equation in the specied interval.
47. x 4
x x,
3
0, 0, 1
1, 2
3 48. sx
1 e x,
x,
0, 1 1, 2
39. f x
1
49. cos x
50. ln x
51 52 (a) Prove that the equation has at least one real root. 40. The gravitational force exerted by the earth on a unit mass at
a distance r from the center of the planet is GMr R3 GM r2 if r if r R R
(b) Use your calculator to nd an interval of length 0.01 that contains a root.
51. cos x
x3
52. ln x
3
2x
Fr
; 53 54 (a) Prove that the equation has at least one real root.
53. 100e
x 100
(b) Use your graphing device to nd the root correct to three decimal places. 0.01x 2
54. arctan x
where M is the mass of the earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r ?
1
x
130
||||
C H A P T E R 2 L I M I T S A N D D E R I VAT I V E S
55. Prove that f is continuous at a if and only if
hl0
62. If a and b are positive numbers, prove that the equation
lim f a
h
fa x3
a 2x 2
1
x3
b x
2
0
56. To prove that sine is continuous, we need to show that
lim x l a sin x sin a for every real number a. By Exercise 55 an equivalent statement is that
hl0
has at least one solution in the interval
63. Show that the function
1, 1 .
lim sin a
h
sin a
fx is continuous on ,
x 4 sin 1 x 0 .
if x if x
0 0
Use (6) to show that this is true.
57. Prove that cosine is a continuous function. 58. (a) Prove Theorem 4, part 3.
64. (a) Show that the absolute value function F x
(b) Prove Theorem 4, part 5.
59. For what values of x is f continuous?
fx
0 1
if x is rational if x is irrational
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, nd a counterexample. usual path to the top of the mountain, arriving at 7:00 P M. The following morning, he starts at 7:00 AM 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.
65. A Tibetan monk leaves the monastery at 7:00 AM and takes his
60. For what values of x is t continuous?
tx
0 x
if x is rational if x is irrational
61. Is there a number that is exactly 1 more than its cube?
2.6
L I M I T S AT I N F I N I T Y ; H O R I Z O N TA L A S Y M P TOT E S In Sections 2.2 and 2.4 we investigated innite limits and vertical asymptotes. There we let x approach a number and the result was that the values of y became arbitrarily large (positive or negative). In this section we let x become arbitrarily large (positive or negative) and see what happens to y. Lets begin by investigating the behavior of the function f dened by x2 1 x2 1 as x becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of f has been drawn by a computer in Figure 1. fx
y
x 0 1 2 3 4 5 10 50 100 1000
fx 1 0 0.600000 0.800000 0.882353 0.923077 0.980198 0.999200 0.999800 0.999998
y=1
0
1
y=
FIGURE 1
-1 +1
x
As x grows larger and larger you can see that the values of f x get closer and closer to 1. In fact, it seems that we can make the values of f x as close as we like to 1 by taking x sufciently large. This situation is expressed symbolically by writing lim x2 x2 1 1 1
xl