654 Appendix C • Answers & Hints for Selected Exercises
- By Parts (a) and (b), lim [f(x)-g(x)] = lim [f(x)+(-l)g(x)] = lim f(x)+
x--+xo x--+xo x--+xo
lim (-l)g(x) = lim f(x) + (-1) lim g(x) = lim f(x) - lim g(x).
x--+xo x-+xo x--+-xo x--+xo x--+xo
lim f(x) lim J(x)
- ( d) lim f ( X) x-+xo ,x-+,..-xo __,,..--~
x-+xo f(x) + 2g(x) lim [f(x) + 2g(x)] lim f(x) + lim [2g(x)]
X--+Xo X-+Xo X--+Xo
lim f(x)
x--+xo^2 -^1
lim f(x) + 2 lim g(x) 2+2·3 4
X--+Xo X--+Xo
- (a) 4 (c)^193 (e) ~ (g) 3 (i) 3/2 (k) 2 fo (m) 3a^2 (o) n
{
- For (a),(b),(c), use f(x) =^1 if. x ->^0 } , g(x) = { -1 if. x > -^0 }.
-1 1f x < 0 1 1f x < 0
- Suppose lim f(x) =Land lim g(x) = M. Let {xn} be any sequence in
X-+Xo X--+Xo
T>(f) n T>(g) - { xo} converging to x 0. By the sequential criterion for limits of
functions, f (xn) ---+ L and g(xn) ---+ M. By the algebra of limits of sequences,
f (xn) + g(xn) ---+ L + M. Therefore, by the sequential criterion for limits of
functions, lim [f(x) + g(x)] = L + M.
x--+xo
13. Done like Exercise 11.
15. Vx =f. xo, f(x) = f((x)) g(x). Thus, lim f(x) = lim f((x)) · lim g(x) = 0.
g x X-+Xo X--+Xo g x X--+Xo
17. Revise the proof of (a) by changing inequalities, etc.
19. Let c > 0. Then :J 81, B > 0 3 0 < Ix - OI < 81 ::::} lg(x)I < B and :J
82 > 0 3 0 < Ix - OI < 82::::} lf(x)I < c/B. Let 8 = min{81,82}. Then 0 <
Ix - OI < 8::::} l.f(x)g(x)I = lf(x)llg(x)I < (c/B)B = c. :. lim f(x)g(x) = 0.
X--+Xo
- Suppose lim f(x) < lim g(x). Let h(x) = g(x) -f(x). Then lim h(x) =
X--+Xo X--+Xo X--+Xo
lim (g(x) - J(x)) = lim g(x) - lim J(x) > 0. So, by Ex. 20, :J 8 > 0 3 0 <
x--+xo x--+xo x--+xo
Ix - OI < 8::::} h(x) > O; i.e., g(x) - f(x) > O; i.e., f(x) < g(x).
- If 0 < lx-xol < 8::::} lf(x)-LI < c then 0 < lx-(xo +p)I = l(x-p)-xol <
8::::} l(f(x - p) - LI < c::::} l(f(x) - LI < c. The argument is reversible.
EXERCISE SET 4.3
- (a),(b) f(x 0 ) = -1 and f(xci) = 1, so lim f(x) does not exist.
x--+xo
(c),(f) f(x 0 ) = f(xci) = 0, so lim f(x) = 0.
X--+XQ
(d) f(x 0 ) does not exist; f(xci) = O; lim f(x) = 0.
X--+Xo
