假设f(x,y)=x2yx2+y2(x2+y2≠0)0(x2+y2=0),试证明:f(x,y)在(0,0)连续,且偏导数存在,但此点

2025-06-24 08:48:53
推荐回答(1个)
回答1:

证明:设x=rcosθ,y=rsinθ,则

lim
(x,y)→(0,0)
f(x,y)=
lim
r→0
r3cosθsinθ
r2
lim
r→0
rsinθcosθ

而sinθcosθ是有界函数
lim
(x,y)→(0,0)
f(x,y)
=0=f(0,0)
故f(x,y)在(0,0)连续
f
(0,0)=
lim
△x→0
f(△x,0)?f(0,0)
△x
lim
△x→0
0
△x
=0

f
(0,0)=
lim
△y→0
f(0,△y)?f(0,0)
△y
lim
△y→0
0
△y
=0

而f(x,y)在(0,0)处的全增量为
△f(0,0)=
x2△y
x2+△y2

△f(0,0)?(
f
(0,0)△x+
f
(0,0)△y)=
x2△y
x2+△y2

lim
ρ→0
△f(0,0)?(
f
(0,0)△x+
f
(0,0)△y)
x2+△y2
=
lim
ρ→0
x2△y
(△x2+△y2)