1549901369-Elements_of_Real_Analysis__Denlinger_

228 Chapter 5 • Continuous Functions

Proof. Consider the sequence { ~}. Observe that ~ -t 0 and sgn ( ~)

1 -t 1 -=f. sgn(O). That is , sgn (~) ft sgn(O). Thus, by Corollary 5.1.4 above,

sgn is discontinuous at 0. 0

Definition 5.1.6 A function f: V(f) -t ffi. is continuous on a set A~ V(f)
if it is continuous at every point of A. If f is continuous on V(f) we say that
f is continuous everywhere on its domain, or simply, f : V(f) -t ffi. is
continuous. If f : ffi.-tffi. is continuous on JR, we say t hat f is continuous
everywhere.

Theorem 5.1.7 Polynomial functions are continuous everywhere.

Proof. Exercise 7. •

Theorem 5.1.8 A rational function R(x) = ~~:~, where p(x) and q(x) are

polynomials, is continuous everywhere on its domain [i.e., at every real number

xo for which q( xo) -I OJ.

Proof. Exercise 8. •

Examples 5.1.9 (a) The absolute value function f(x) = JxJ is continuous
everywhere.
(b) The square root function f ( x) = .jX is continuous everywhere on its
domain [O, + oo).

Proof. Exercise 9. 0

Examples 5.1.10 (a) The function f(x)

x = 3 but

2x^2 - 18

is not continuous at

x-3

{

2x2 - 18 if x -I 3 }

(b) The function g(x) = x - 3 is continuous at x = 3.

12 if x = 3

Proof. (a) Since f(3) does not exist, f is not continuous at x = 3.
(b) In Example 4.1.4, we saw that lim g(x) = 12. By definition, g(3) = 12.
X->3
Thus, lim g(x) = g(3), and therefore, g is continuous at x = 3. D
X->3

Example 5.1.11 (A Function That Is Continuous Nowhere)

The Di'ri'chlet f unc t' ion f ( ) x = { 1 if. x.. is rational. } is. d ' IScontmuous.
0 if x is irrational
everywhere.