证明:函数-|||-f(x,y)= ((x)^2+(y)^2)sin dfrac (1)(sqrt {{x)^2+(y)^2}}, ^2+(y)^2neq 0,-|||-0, ^2+(y)^2=0-|||-在点(0,0)连续且偏导数存在,但偏导数在点(0,0)不连续,而f在点(0,0)可微.

函数f(x,y)= dfrac (xy)({x)^2+(y)^2},(x)^2+(y)^2neq 0-|||-0, ^2+(y)^2=0在点（0，0）处（）。

设 f(x,y)= sin dfrac (1)({x)^2+(y)^2} , ^2+(y)^2neq 0,-|||-0, ^2+(y)^2=0,-|||-考察函

设f(x,y)= ^4+{y)^2},(x)^2+(y)^2neq 0 0,(x)^2+(y)^2=0 .处是否连续？设,试问在点处是否连续？

有关函数f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-)在点（0，0）的性态，下列说法

讨论下列函数的连续性:-|||-^2+(y)^2neq 0,-|||-f(x,y)= ) (y)^2ln ((x)^2+(y)^2) 0,=0;

( A ) = (x,y,z)|{x)^2+(y)^2+(z)^2=(a)^2,zgeqslant 0} ( B ) = (x,y,z)|{x)^2+(y)^

设函数y=f(x)由方程(y)^2+(y)^2ln x-4=0所确定,则(y)^2+(y)^2ln x-4=0= ( )(y)^2+(y)^2ln x-4=0(

设函数 (x,y)=1-dfrac (cos sqrt {{x)^2+(y)^2}}(tan ({x)^2+(y)^2)} ,则当定设函数 (x,y)=1-df

设 (x,y)=dfrac ({x)^2+(y)^2}({e)^xy+xysqrt ({x)^2+(y)^2}} ,则 (f)_(x)(1,0)= __ _.

已知Σ为锥面=sqrt ({x)^2+(y)^2}在柱体=sqrt ({x)^2+(y)^2}内的部分,则曲面积分=sqrt ({x)^2+(y)^2}