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，下列说法正确的有（）个.(1)由于f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-)（当f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-)时），所以由夹逼准则知f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-)，从而有f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-)；(2)由于f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-)，所以连续；(3) f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-),f(x,y)= dfrac (xy)(sqrt {{x)^2+(y)^2}},(x)^2+(y)^2neq 0-|||-),所以偏导数存在A.0个B.1个C.2个D.3个

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（）。

证明:函数-|||-f(x,y)= ((x)^2+(y)^2)sin dfrac (1)(sqrt {{x)^2+(y)^2}}, ^2+(y)^2neq 0,

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？

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;

8.函数f(x,y)=}(xy)/(x^2)+y^(2),x^2+y^2neq0,0,x^2+y^2=0,在(0,0)点（）A. 连续，偏导函数都存在B. 不连

2【单选题】设f(x,y)=}(x^2+y^2)sin(1)/(sqrt(x^2)+y^(2)),(x,y)neq(0,0)0,(x,y)=(0,0)f(x,y

[题目]已知函数f(x,y)在点(0,0)的某个邻域内连-|||-续,且 lim _(xarrow 0)dfrac (f(x,y)-xy)({({x)^2+(y

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

10.判断题-|||-二元函数-|||-.f(x,y)= ^2+{y)^2}}(x,y)neq (0,0) 0,(x,y)=(0,0) .-|||-点(0