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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?

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  • $\begingroup$ Actually the function isn't even continuous at $(0,0)$. Check for example two paths $(x,0)\to (0,0)$ and $(x,x)\to (0,0)$ $\endgroup$ – Mostafa Ayaz Mar 24 '19 at 7:54
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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 $(0,0)$.

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  • $\begingroup$ I forgot to put in the full question, my bad. But i need to make sure if $f_{yx}$ (0,0) exists or not. $\endgroup$ – crown Mar 24 '19 at 8:00
  • $\begingroup$ You can see it is non-existent from your calculations above $\endgroup$ – Tojrah Mar 24 '19 at 8:03

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