because we obtained via sloppy estimates.
As a matter of fact, in this case must depend
on x0 as well as ; see Example 3. Figure 19.1
shows how a fixed requires smaller and
smaller as x0 approaches 0. [In the figure, 1
signifies a that works for x1 and ,
limxs f(x) = L if and only if for each there
exists such that implies . (1) For finite limits L,
the first and last blanks are filled in by > 0
and |f(x)L| < . For L = +, the first and last
blanks are filled in by M > 0 and f(x) > M,
while for L = they ar
b], (1) implies |fj (t) fj (t0)| d(t), (t0)
for j = 1,., k, (2) and d(t), (t0) k
maxcfw_|fj (t) fj (t0)| : j = 1, 2,.,k. (3) Suppose
is continuous on [a, b], and consider some j
in cfw_1, 2,.,k. To show fj is continuous at t0 in
[a, b], given > 0, select
for x > 0 and f(x) = 0 for x 0, x0 = 0; (b) g(x) =
sin( 1 x) for x = 0 and g(0) = 0, x0 = 0; (c) sgn(x)
= 1 for x < 0, sgn(x) = 1 for x > 0, and sgn(0) =
0, x0 = 0. The function sgn is called the signum
function; note sgn(x) = x |x| for x = 0. 17.11 Let
f
f(y)| 1. (1) [Actually, for this function, (1) in
Definition 19.1 fails for all > 0.] To show (1) it
suffices to take y = x + 2 and arrange for
f(x) f x + 2 1. (2) [The motivation for
this maneuver is to go from two variables, x
and y, in (1) to one varia
computational software programs provide
formulas like ! x2 cos x dx = 2x cos x + (x2 2)
sin x + C. It is straightforward to show the
derivative of each function 2x cos x+ (x2 2)
sin x + C is in fact x2 cos x. Corollary 29.5
shows that these are the only a
and Theorem 17.2. It would be reasonable to
view f(a) as the limit of the values f(x), for x
near a, and to write limxa f(x) = f(a). In this
section we formalize this notion. This section
is needed for our careful study of derivatives in
Chap. 5, but it m
= 0 and f(0) = 0. (a) Observe f is continuous on
R; see Exercises 17.3(f) and 17.9(c). (b) Why is f
uniformly continuous on any bounded subset
of R? (c) Is f uniformly continuous on R? 19.10
Repeat Exercise 19.9 for the function g where
g(x) = x2 sin( 1 x
Let f be a continuous strictly increasing
function on some interval I. Then f(I) is an
interval J by Corollary 18.3 and f 1 represents
a function with domain J. The function f 1 is a
continuous strictly increasing function on J.
Proof The function f 1 is
continuous function f: R R looks like a curve,
and it is! It is the curve for the path (t)=(t,
f(t). 166 3. Continuity FIGURE 21.1 FIGURE
21.2 Curves in R3 can be quite exotic. For
example, the curve for the path h(t) = (cost,sin
t, t 4 ) is a helix. See
1 (x2)3 for x = 2. Then limx f(x) =
limx f(x) = 0, limx2+ f(x)=+ and
limx2 f(x) = . To verify limx f(x) = 0,
we consider a sequence (xn) such that
limn xn = + and show limn f(xn) = 0.
This will show limxS f(x) = 0 for S = (2,),
for example. Exercise 9.11
Theorem 18.1 is false if the closed interval [a,
b] is replaced by an open interval. For
example, f(x) = 1 x is continuous but
unbounded on (0, 1). The function x2 is
continuous and bounded on (1, 1), but it does
not have a maximum value on (1, 1). 18.2
I
limit exists for the other choice of S and these
limits are identical. Their common value is
what we write as limxa+ f(x). 20.20 Let f1
and f2 be functions such that limxaS
f1(x)=+ and such that the limit L2 = limxaS
f2(x) exists. (a) Prove limxaS (f1 + f
(a, b) and x2 (b, c) such that f(x1) = f(x2) = y.
This contradicts the one-to-one property of f.
Now select any a0 < b0 in I and suppose, say,
that f(a0) < f(b0). We will show f is strictly
increasing on I. By (1) we have f(x) < f(a0) for
x< b0], f(a0) <
(f(sn) is also a Cauchy sequence. Example 6
We show f(x) = 1 x2 is not uniformly
continuous on (0, 1). Let sn = 1 n for n N.
Then (sn) is obviously a Cauchy sequence in (0,
1). Since f(sn) = n2, (f(sn) is not a Cauchy
sequence. Therefore f cannot be unifo
S is continuous on S if and only if f 1(U) is
an open subset of S for every open subset U of
S. (1) 168 3. Continuity Recall f 1(U) = cfw_s S
: f(s) U. Proof Suppose f is continuous on S.
Let U be an open subset of S, and consider s0
f 1(U). We need to
obtain the condition (1) in Discussion 20.9
equivalent to limx f(x) = L. In the same
way, the limits limx f(x) and limxa f(x)
will equal L [L finite] if and only if for each > 0
there exists < . (2) Obvious changes are
needed if L is infinite. Exercises 2
hence g(x0) < g(x0) + , and hence |g(x)
g(x0)| < . Now if we put = mincfw_x2 x0, x0
x1, then |x x0| < implies x1 <x< . Theorem
18.5 provides a partial converse to the
Intermediate Value theorem, since it tells us
that a strictly increasing function with
' limn f2(xn) ( = L1L2. Thus (2) in
Definition 20.1 holds for f1 +f2 and f1f2, so
that (i) and (ii) hold. Likewise (iii) follows by an
application of Theorem 9.6. Some of the
infinite variations of Theorem 20.4 appear in
Exercise 20.20. The next theorem
every sequence (xn) in dom(f) \ cfw_x0
converging to x0, we have lim f(xn) = f(x0).
17.16 The postage-stamp function P is defined
by P(x) = A for 0 x < 1 and P(x) = A + Bn for n
x n. By the BolzanoWeierstrass Theorem
11.5, (xn) has a subsequence (xnk ) t
|x2y2| = |xy|x+y|. Since |x + y| 14 for x,
y in [7, 7], we have |f(x) f(y)| 14|x y| for
x, y [7, 7]. Thus if = 14 , then x, y [7,
7] and |x y| < imply |f(x) f(y)| < . We
have shown that f is uniformly continuous on
[7, 7]. A similar proof would work for f
3. Continuity If we can show y+x x2y2 is
bounded on [a,) by a constant M, then we
will take = M . But we have y + x x2y2 = 1 x2y
+ 1 xy2 1 a3 + 1 a3 = 2 a3 , so we set = a3 2 .
It is now straightforward to verify (1). In fact, x
a, y a and |x y| < imply
x0| < imply |f(x) f(x0)| < . (*) The choice
of depends on > 0 and on the point x0 in S.
140 3. Continuity Example 1 We verify (*) for
the function f(x) = 1 x2 on (0,). Let x0 > 0 and
> 0. We need to show |f(x) f(x0)| < for |x
x0| sufficiently small. Not
differentiable on I and if fis bounded on I ,
then f is uniformly continuous on I. Proof For
this proof we need the Mean Value theorem,
which can be found in most calculus texts or
later in this book [Theorem 29.3]. Let M be a
bound for fon I so that |f (
clearly f is uniformly continuous on (a, b).
Suppose now that f is uniformly continuous on
(a, b). We need to define f(a) and f(b) so that
the extended function will be continuous. 19.
Uniform Continuity 149 It suffices for us to
deal with f(a). We make t
using appropriate theorems or Exercise
19.4(a). (a) tan x on [0, 4 ], (b) tan x on [0,
2 ), (c) 1 x sin2 x on (0, ], (d) 1 x3 on (0, 3),
(e) 1 x3 on (3, ), (f) 1 x3 on (4, ). 19.6 (a)
Let f(x) = x for x 0. Show f is unbounded on
(0, 1] but f is neverthel
f(y)| < . (1) We will say f is uniformly
continuous if f is uniformly continuous on
dom(f). 19. Uniform Continuity 141 FIGURE
19.1 Note that if a function is uniformly
continuous on its domain, then it is continuous
on its domain. This should be obvious;
= inf # f(x) : x $ i n , i+1 n %& . Then the sum
of the areas of the rectangles in Fig. 19.2a
equals Un = 1 n n 1 i=0 Mi,n and the sum of
the areas of the rectangles in Fig. 19.2b equals
Ln = 1 n n 1 i=0 mi,n. The function f would
turn out to be Riemann i
f(yn)| . By the Bolzano-Weierstrass Theorem
11.5, a subsequence (xnk ) of (xn) converges.
Moreover, if x0 = limk xnk , then x0
belongs to [a, b]; see Exercise 8.9. Clearly we
also have x0 = limk ynk . Since f is
continuous at x0, we have f(x0) = limk
f(xn
limxa f(x). In fact, f(a) = limxa f(x) if and
only if f is defined on an open interval
containing a and f is continuous at a. (b) For a
R and a function f we write limxa+ f(x) = L
provided limxaS f(x) = L for some open
interval S = (a, b). limxa+ f(x) is