MAT208 - Exerc´ıcios LISTA 4 - 2011
1. Determine o limite se existir, ou mostre que o limite n˜ao existe (justifique sua resposta):
(a) lim
(x,y)→(5,−2)(x5 + 4x3y−5xy2) (b) lim
(x,y)→(0,0)
3x2−y2+ 5 x2+y2+ 2 (c) lim
(x,y)→(0,0)
x3+xy2
x2+y2 (d) lim
(x,y)→(1,1)
x−y
x3−y3 (e) lim
(x,y)→(0,0)
x3y3 2x2+y2 (f) lim
(x,y)→(0,0)
8x2y2
x4+y4 (g) lim
(x,y)→(0,0)
8x4y x4+y4 (h) lim
(x,y)→(0,0)
xy3
x2+y6 (i) lim
(x,y)→(0,0)
xy px2+y2 (j) lim
(x,y,z)→(3,0,1)e−xy sen (πz
2 ) (k) lim
(x,y,z)→(0,0,0)
x2+ 2y2+ 3z2 x2+y2+z2 (l) lim
(x,y,z)→(0,0,0)
xyz2
x2+y2 +z2 (m) lim
(x1,x2,x3,x4)→(1,1,1,1)(x12 +x22+x32+x42)
2. Calcule o limite:
(a) lim
(x,y)→(6,3)xycos(x−2y) (b) lim
(x,y)→(π,2π)
cosy+ 1
y−sinx (c) lim
(x,y)→(1,1)ln|x2y2−2|
(d) lim
(x,y)→(1,1)lnx3−y3
x−y (e) lim
(x,y)→(2,2)
x+y−4
√x+y−2 (f) lim
(x,y)→(0,0)
x2+y2 px2+y2+ 1−1 (g) lim
(x,y)→(0,0)
sin 2x
x(y2−1) (h) lim
(x,y)→(0,0)
(xy−2) tany2 y2
3. Determine o maior conjunto no qual a fun¸c˜ao ´e cont´ınua:
(a) f(x, y) = 1
x2−y (b) f(x, y) = ln(2x+ 3y) (c) g(x, y) = arctan(x+√ y)
(d) h(x, y) = sin 1
xy (e) f(x, y) = x+y
2 + cosx (f) g(x, y) = x2+y2 x2−3x+ 2 (g) lim
(x,y)→(0,0)
x3y3 2x2+y2
1
(h) f(x, y) =
( x3y3
2x2+y2 se(x, y)6= (0,0) 1 se(x, y) = (0,0)
(i)f(x, y) =
( x3y3
2x2+y2 se(x, y)6= (0,0) 0 se(x, y) = (0,0) (j) g(x, y, z) = xyz
x2+y2−z (k)f(x, y, z) = xyz px2+y2−z (l)f(x, y, z) = ln(x2+y2+z2) + z2
x2+y2 (m) f(x, y, z, t) = x2 +y2 x+y+z−t 4. Calcule o vetor gradiente de f em cada ponto (x, y) se
(a) f(x, y) =x5+ 3x3y2+ 3xy4 (b) f(x, y) = (xy+ 1)2 (c) f(x, y) =e−xsen(x+y)
5. Mostre que as derivadas parciais de f(x, y) =
xy
x2+y2 se(x, y)6= (0,0) 0 se(x, y) = (0,0)
existem em (0,0) e calcule-as. Mas mostre tambem que f n˜ao ´e cont´ınua em (0,0).
Algumas respostas 1. (a) 2025 (b) 52 (c) 0 (d) 13 e) 0 (f) n˜ao existe
(g) 0 (h) n˜ao existe i) 0 (j) 1 (k) n˜ao existe (l) 0 (m) 4 2. (a) 18 (b) π1 (c) 0 (d) ln 3 (e) 4 (f) 2 (g) −2 (h) −2 3. (a) {(x, y)|y 6=x2} (b) {(x, y)|y >−23x} (c) {(x, y)|y >0}
(d) {(x, y)|x6= 0, y 6= 0} (e) R2 (f) {(x, y)|x6= 1, x6= 2} (g) R2 − {(0,0)} (h) R2− {(0,0)}
(i)R2 (j) {(x, y, z)∈R3 | z 6=x2+y2}
(k) {(x, y, z)| z < x2+y2}(abaixo do paraboloide z =x2+y2)
(l){(x, y, z) | x2+y2 6= 0}=R3−eixo 0z (m) {(x, y, z, t)∈R4 | x+y+z 6=t}
4. (b) ∇(x, y) = (2y(xy~ + 1),2x(xy+ 1)) 5. ∂f∂x(0,0) = 0, ∂f∂y(0,0) = 0
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