Answers & Hints for Selected Exercises 667. f(x) - f(O)..
(b) hm
0
= hm xr-l sm ~. By the above argument,
x->O+ X - x->O+
(i) If r > 1, lim xr-l sin~= 0, so f~(O) = 0.
x->O+
(ii) If r ~ 1, lim xr-l sin~ does not exist, so f~ (0) does not exist.
x->O+- f'(xo) exists~ 3L E JR. 3 lim f(x) - f(xo) = L
x->xo x - Xo
~ 3L E JR. 3 lim f(x) - f(xo) =Land lim f(x) - f(xo) = L
x->x() X - Xo x->xt X -Xo
~ f '_ ( Xo) and f ~ ( Xo) exist and are equal. - Let f(x) = ~ if x f 0, Oifx = 0. Then f'(x) = 0 ifx f 0, so lim f'(x) =
x->0-
0 = lim J'(x). But lim J(x)-f(O) = lim .!. and lim f(x)-f(O) = lim .!. so
x->O+ x->O- x-O x->O-x x->O+ x-O x->O+ x'
f'_ ( xo) and f~ ( xo) do not exist. - Assume the hypotheses. Then 3L E JR. 3 lim f'(x) = lim f'(x) = L. By
x->O- x->O+
Thm. 6.1.14, f'_(xo) = f~(xo) = L. By Thm. 6.1.13, f'(xo) exists and= L. - (a) Suppose 3a > 1andM>0 3 Vx,y EI, lf(x)-f(y)I ~ Mlx-yl°'· Let
xo EI. Then Vx f xo in I, lf(x) - f(xo)I ~Mix - xol°', so I f(x) - f(xo) I ~
x-xo
Mix - xol°'-^1. Since a> 1, lim (x - xo)°'-^1 = 0 since the power function is
X--+Xo
continuous.:. lim f(x) - f(xo) = 0.
x->xo x - xo
(b) Take f(x) = lxl on I= (0, 1).
EXERCISE SET 6.2- By Parts (a) and (b), (f - g)'(xo) = (f + (-g))'(xo) = f'(xo) + (-g)'(xo) =
f'(xo) -g'(xo).
d [ 1 J f(x)fxl - lfxf(x) -f'(x)
- By Thm. 6.2.2, dx f(x) = J2(x) = J2(x).
f(x) - f(-xo)
- (a) If f is odd and differentiable at xo, then f'(-xo) = lim ( ) =
X->-Xo X - -Xo
lim -f(-x) + f(xo) = lim f(-x) - f(xo) = lim f(u) - f(xo) = f'(xo).
x->-xo -(-x)+xo -x->xo -X-Xo u->xo u-xo
(b) If f is even and differentiable at x 0 , then f'(-x 0 ) = lim f(x) -/(-~o) =
X->-Xo X - -Xo1. f(-x) - f(xo)
1
. f(-x) - f(xo)
1
. f(u) - f(xo)
Im - Im - Im
x->-xo x + x 0 -x->xo - x - x 0 u->xo u - xo
- f'(xo).