Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 14.2 LIMITS AND CONTINUITY 0 193
11. j(x , y) = (y 2 sin 2 x)j(x 4 + y 4 ). On the x -axis, f(x, 0) = 0 for x '# 0, so f(x, y) --+ 0 as (x, y) --+ (0, 0) along the
. x 2 sin 2 x sin 2 x 1 ( sinx) 2
x-axis. Approaching (0, 0) along the hne y = x, f(x, x) = x 4 + x 4
=
2 x 2 = 2 --;-- for x :/: 0 and
lim sin x = 1, so f(x, y)--+ ~·S ince f has two different limits along two different lines, the limit does not exist.
:c- o x · ·
13. f(x, y) = ~· We can see that the limit along any line through (0, 0) is 0, as well as along other paths through
xz +yz
(0, O) such as x = y 2 andy= x 2 . So we suspect that the limit exists and equals 0; we use the Squeeze Theorem to prove our
assertion. 0 ~ I ~~~ lxl since lvl ~ )x 2 + y 2 , and lxl --+ 0 as (x, y)--+ (0, 0). So lim f(x, y) = o:
x2 + y2 . (x,y)-(o,o)
15. Let f(x, y) =
x2yeY
+ 4y
X
4 2
. Then f(x, 0) = 0 for x '# 0, so f(x, y) --+ 0 as (x, y)--+ (0, 0) along the x -axis. Approaching
2 2 ::r2 4 ::c2 a:2
(0, 0) along they-axis or the line y = x also gives a limit ofO. But f (x, x 2 ) = x:: 4
tx 2
)2 = x 5
: 4
= T for x '# 0, so
f(x, y) -+ e 0 /5 = t as (x,y)--+ (0, 0) along theparabola y = x 2 . .Thus the limit doesn't exist.
19. eY 2 is a composition of continuous functions and hence continuous. xz is a continl.!ous function and tan t is continuous for
t '# ~ + mr (nan integer), so the composition tan(a