Solução_Calculo_Stewart_6e

F.
176 ¤ CHAPTER 15 PARTIAL DERIVATIVES ET CHAPTER 14
TX.10
x 3 + y 3
39. lim
(x,y)→(0,0) x 2 + y = lim (r cos θ) 3 +(r sin θ) 3
= lim
2 r→0 + r 2
r→0 +(r cos3 θ + r sin 3 θ)=0
41. lim
(x,y)→(0,0)
e −x2 −y 2 − 1 e −r2 − 1 e −r2 (−2r)
= lim = lim
x 2 + y 2 r→0 + r 2 r→0 + 2r
[using l’Hospital’s Rule]
= lim
r→0 + −e−r2 = −e 0 = −1
⎧
⎨ sin(xy)
if (x, y) 6= (0, 0)
43. f(x, y) = xy
⎩
1 if (x, y) =(0, 0)
From the graph, it appears that f is continuous everywhere. We know
xy is continuous on R 2 and sin t is continuous everywhere, so
sin(xy) is continuous on R 2 and sin(xy) is continuous on R 2
xy
except possibly where xy =0. To show that f is continuous at those points, consider any point (a, b) in R 2 where ab =0.
Because xy is continuous, xy → ab =0as (x, y) → (a, b). Ifwelett = xy, thent → 0 as (x, y) → (a, b) and
sin(xy) sin(t)
lim
=lim
(x,y)→(a,b) xy t→0 t
on R 2 .
=1by Equation 3.4.2 [ET 3.3.2]. Thus
lim
(x,y)→(a,b)
f(x, y) =f(a, b) and f is continuous
45. Since |x − a| 2 = |x| 2 + |a| 2 − 2 |x||a| cos θ ≥ |x| 2 + |a| 2 − 2 |x||a| =(|x| − |a|) 2 ,wehave |x| − |a|
≤ |x − a|. Let
>0 be given and set δ = . Thenif0 < |x − a|