Question:$f(x,y)$ = $\lbrace$$\frac{2xy}{x^2+y^2}$ if $(x,y)$ $\neq (0,0$), $0 $ otherwise$\rbrace$
Does $f_{yx}$$(0,0) exist?$
Attempt:
$f_x$=$\frac{2y(y^2-x^2)}{(x^2+y^2)^2}$
$f_y$=$\frac{2x(1-2y^2)}{(x^2+y^2)^2}$
$f_{xy}$=$\lim_{h\to 0}$$\frac{f_x(0,0+h)-f_x(0,0)}{h}$=$\lim_{h\to 0}$$\frac{2}{h^2}$ $\to$ +infinity
$f_{yx}$=$\lim_{h\to 0}$$\frac{f_y(0+h,0)-f_y(0,0)}{h}$=$\lim_{h\to 0}$$\frac{2}{h^4}$ $\to$ +infinity
So the limit does not exist. Is this correct? If both limit goes to infinity it means limit does not exist right?
Yes,the limit does not exist. Otherwise also suppose you approach $(0,0)$ along $y=mx$ so putting y=mx gives, $f(x,mx)=\frac {2m}{1+m^2}$. The value the function tends to at $(0,0)$ varies with $m$. So the limit is non-existent at $(0,0)$.