Answers & Hints for Selected Exercises 673G'(x) = -g'(-x). Consequently,
(a) F, G: - I -t IR where -I is an interval with left end-point -x 0 ;
(b) F, G are differentiable on -I;
(c) \:/x E -I, -x EI so g(-x)g'(-x) # O; i.e., G(x)G'(x) # O;
(d) lim F(x) = lim f(-x) = lim f(x) = 0 (using Ex. 2);
x-+-xci x----+-xci x-+xQ
Similarly, lim G(x) = O;
X-+-xt
(e) ll·m G'(x) F'(x) --^11 ·m -f'(-x) -g'(-x) --^11 ·m + g'(J'(-x) -x) = 11·m _ J'(x) g'(x) = L.
X--+-xt X-+-xt X--+-Xo X--+Xo.. · B y C ase^1 , l" lm + G(x) F(x) -- L , so b E y x.^2 , 1. im F(-x) G(-x) -- L · , · Le., 1. lm fJ=l g(x) -- L.
X-+-Xo X--+Xo X----+Xo
5. See how Exercise 3 was done.
7. Modify the proof of Case 2 by changing "> M + 1" to "< -M - 1" and
"> M" to"< -M" etc.
9. To prove Cases 7- 9, put together Cases 1 & 4, Cases 2 & 5, and Cases 3 & 6.
11. To prove Cases 13- 15, modify the proofs of Cases 10- 12 in obvious ways.
12. (a) +oo (c) 0 (e) -oo (g) 1
(b) lim e; = lim e; = O; (L'Hopital's rule is optional here.)
X-+- 00 X-+-00
(d) lim secx = lim secxtanx = lim xsecxtanx = 0.
x->O+ In x x->O+ l / x x->O+
( f) lim 3+4 sec x = lim 4 sec x2tan x = 4. lim tan x = 4. lim sin X = 4.
x->(~)- 2+tanx x->(~)- sec x x->(~)-secx x->(~)-
Without L'Hopital's rule 'x->(~)-lim 2cosx+smx^3 cos x-i:^4 = 1 1 = 4.
(h) x_,lim l/(x-2) - rm -(x-2)-^2 = - lim _l_ = -()().
2 + ln(x-2) - x2. 2 + (x-2)(^1) x_,
2 + x-2
- (b) 1 (d) 1 (e) 1 /e (g) e (h) 1/2
(a) lim (secx - tanx) = lim l-sinx = lim -c?sx = Q = O.
x->(~)+ x->(~)+ cosx x->(~)+ - smx 1
() r^1 (. ) - r ln(sinx) - r cosx/sinx - r x
(^2) cosx
C x2.1g+ X n sm X - x2.1g+ I/x - x2.1g+ -l/x2 - - x2.1g+ SinX
= _ lim - x^2 sin x+2x cos x = 0.
x->O+ cosx
(f) T ake logarithm. ln [ lim X-+00 (1 - l/x^2 r ] = X-+00 lim ln (1 - l/x^2 r =
lim xln (1-1/x^2 )
X->00
- r ln(l-1/x^2 ) - r ln(x^2 -1)-lnx^2 - r ~-~ - r 2x^3 -2x(x^2 -1)
- x.2..~ 1/x - x.2..~ 1/x - x.2..~ -1/x^2 - x.2..~ -(x^2 -1)
= X----+00 lim -< X^2 t_ 1 l = X--+00 lim -~ X = o. .·. X-+00 lim (1 - 1 /x^2 r = 1.
- (a) We use mathematical induction to prove that \:/k E N, 3 polynomial
Pk(x) with constant term 0 such that \:/x > 0, J(kl(x) =Pk(~) e-l/x.