Does this partial derivative exist at (0,0)?

Question

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?

Answer 1

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 $(0,0)$.