658 Appendix C • Answers & Hints for Selected Exercises
:::::} l4x + 31 Ix - 21 < 15 1 e 5 :::::} l(4x^2 - 5x - 3) - 31 < c
:::::} ll(x) -1(2)1 < c.
- Let p(x) be a polynomial. Every x 0 is a cluster point of V(f) = (- 00,00),
and lim p(x) = p(x 0 ), by Thm. 4.2.13. Thus, by note (2) following Defn. 5.1.1,
X-+Xo
p( x) is continuous at x 0.
- Let l(x) = lxl and Xo E R For c > 0, choose 8 = €. Then Ix - xol < 8 :::::}
ll(x) - l(xol = I lxl - lxol I :::; Ix - xol < c by the triangle inequality, Thm.
1.2.15 (c).
- (a) 0 is a cluster point of7J(J) = (-00,00). liml(x) does not exist, by
x-->O
(4.1.12). So, by note (2) following Defn. 5.1.1, 1 is not continuous at 0.
(b) 0 is a cluster point of V(g) = (-oo, oo). lim g(x) = 0 = g(O) as shown
X-->0
in (4.2.21). So, by note (2) following Defn. 5.1.1, g is continuous at 0.
- (a) Continuous on (-00,00).
(c) Continuous on (-oo,-3), (-3, 1), and (1,oo).
(e) Continuous on (-oo,-2), (-2, 2), and (2,oo).
(g) Continuous on [O, oo).
(h) Continuous on every interval [ -~ + 2mr, ~ + 2mr] , n E Z.
(j) Continuous everywhere.
- Modify the proof of (b) appropriately.
- Let {xn} be a sequence in 7J(J) n 7J(g) 3 Xn ---+ xo. By the sequen-
tial criterion for continuity (5.1.3), f(xn) ---+ f(xo) and g(xn) ---+ g(xo). By
Ex. 2.2.21, max{f(xn),g(xn)}---+ max{f(xo),g(xo)} and min{f(xn),g(xn)}---+
min{f(x 0 ),g(x 0 )} .. '.Therefore, max{f,g} and min{f, g} are continuous at x 0.
- f ( x) = 1 if x is rational, -1 if x is irrational.
2 1. Use f(x) = { 1.if x.is.rati~nal } ' g(x) = { O.if x.is.rati~nal }·
0 1f x 1s irrational 1 if x is irrational
- (a) f-^1 (-00, a)= LJ:=i f-^1 (a-n, a). Each f-^1 (a-n, a) is open by Ex. 22,
and the union of open sets is open. Similarly, f-^1 (a, +oo) = LJ:= 1 f-^1 (a, a+ n)
and each 1-^1 (a,a+n) is open.
(b) f-^1 (-oo,a] =IR-f-^1 (a, + oo); 1-^1 [a,+oo) =IR-1-^1 (-oo,a).
(c) f-^1 (a) =JR - u-^1 (-oo, a) u f-^1 (a, + oo))' so it is closed.
