Question

证明函数f(x,y)={(x的平方+y的平方)乘cos1/(x平方+y的平方),x平方+y的平方#0时; 0,x平方+y

Answer 1

lim(x->0,y->0)f(x,y)=0=f(0,0),所以：在点（0,0）连续

fx(0,0)=lim(x->0){[x^2*cos(1/x^2)]-f(0,0)}/x

=lim(x->0)x*cos(1/x^2)=0

fy(0,0)=lim(y->0){[y^2*cos(1/y^2)]-f(0,0)}/y

=lim(y->0)y*cos(1/y^2)=0