Why Read This Book? - Index of

226 Chapter 7 Functions of a Real Variable
Notice that the hypothesis condition ofTheorem 7.5.10 does not require an �= a
as Exercise 7.3.6 does. For if any of the an = a, then f(an) = f(a), so that the
sequence 〈f(an)〉 is defined for all n.
Here is an important result we could not address with limits alone. A composition
function g ◦ f is continuous at a point if f and g are continuous at the
right points. To prove this, an arbitrarily chosen ɛ-neighborhood of g[f(a)] produces
a δ1-neighborhood of f(a), and this δ1-neighborhood of f(a) produces a
δ2-neighborhood of a.
EXERCISE 7.5.11 Suppose f is continuous at a and g is continuous at f(a).
Then g ◦ f is continuous at a.
Another way to write what Exercise 7.5.11 says is
lim g(f(x)) = g(lim f(x)) = g(f(lim x) = g(f(a)) (7.18)
x→a x→a x→a
With a little more mathematical machinery than we have up to now, Eq. (7.18)
comes in handy in certain manipulations of limits. If we assume for the moment
that all the functions below are continuous at the points involved, Eq. (7.18) allows
us to write something like
�
lim
x→3
1 + e−x2 �
= lim
x→3 (1 + e−x2 )
�
= 1 + lim e
x→3 −x2
�
=
�
= 1 + e−9 1 + e limx→3(−x 2 )
(7.19)
We could not have a theorem in Section 7.2 that said something like “If f(x) →
L1 as x → a and g(x) → L2 as x → L1, then (g ◦ f)(x)→ L2 as x → a” because
a possible hole in the domain of g at f(a) might cause troublesome gaps in the
domain of g ◦ f around a.
Example 7.5.12 Let f(x) = x cos(1/x) and g(x) = sin x/x, and let a = 0. Now
cos(1/x) is bounded, so that f(x) → 0asx → 0. Also, by the presence of cos(1/x),
f(x) = 0 for infinitely many values of x in every neighborhood of zero. Thus
(g ◦ f)(x)fails to exist at infinitely many points in every neighborhood of zero
because g(0) is not defined. �
If f is discontinuous at a point it could be for one or more of three basic
reasons. If Eq. (7.16) is not satisfied it might be that