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prove that the function defined by $f(x,y) = \frac{|x|\cdot y^2}{|x|+y^2}$ for $(x,y)\neq(0,0)$ and $f(0,0)=0$ is differentiable at $(0,0)$

I proved that the partial derivatives at $(0,0)$ exist (they both equal zero) but I have trouble proving that they are continous. How do I show that this limit $$\lim \limits_{(x,y)\to(0,0)} f_x(x,y) = \lim \limits_{(x,y)\to(0,0)} y^2\cdot \frac{2x+y^2sgn(x)}{\left ( |x|+y^2\right )^2}$$
is equal to $0$? and that's even before I looked into the other partial derivative, which I assume wouldn't be easy either.

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  • $\begingroup$ Choose two sequences, one approaching 0 from left and one from the right. If they coincide... $\endgroup$
    – wueb
    Jul 14, 2017 at 18:39
  • $\begingroup$ Note that $f_x=\dfrac{y^2\textbf{sgn}(x)}{(|x|+y^2)^2}$. $\endgroup$ Jul 14, 2017 at 18:55
  • $\begingroup$ @JohnWaylandBales should be $y^4$ in the numerator, not $y^2$ $\endgroup$ Jul 14, 2017 at 18:58
  • $\begingroup$ @eyeballfrog Correct, it was a typo but too late to correct. Thanks! $\endgroup$ Jul 14, 2017 at 19:01

2 Answers 2

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after $AM-GM$ we have $$\frac{|x|+y^2}{2}\geq |y|\sqrt{|x|}$$ form here we get $$\frac{|x|y^2}{|x|+y^2}\le \frac{|x|y^2}{2|y|\sqrt{|x|}}=\frac{1}{2}|x|^{3/2}|y|->0$$

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  • $\begingroup$ OP: Please reread the definition of differentiability, then explain why your post does not even address the differentiability of $f$. // @upvoter Please explain your vote. $\endgroup$
    – Did
    Jul 16, 2017 at 12:57
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The partial derivatives need not be continuous in order the function is differentiable.

You know that the differential, if it exists, is the zero map. Now what you need is that $$ \lim_{(x,y)\to(0,0)}\frac{1}{\sqrt{x^2+y^2}}\left|\frac{|x|\cdot y^2}{|x|+y^2}\right|=0 $$ On the other hand, $$ \frac{|y|}{\sqrt{x^2+y^2}}\le1 \qquad \frac{|x|}{|x|+y^2}\le1 $$ so $$ \frac{1}{\sqrt{x^2+y^2}}\left|\frac{|x|\cdot y^2}{|x|+y^2}\right| \le |y| $$

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