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Take the function $$ \begin{cases}f(x,y)=(x^2+y^2)/\sin(\sqrt{x^2+y^2}) & \text{when } 0<\lvert (x,y)\rvert<\pi\\ 0 & \text{when } (x,y)=(0,0) \end{cases}$$

I got that this function is differentiable at $(0,0)$. I calculated the partial derivatives at $(0,0)$ of which both were $0$ and trying to show the definition of differentiability is not satisfied but I keep getting that it is. Plz help.

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    $\begingroup$ The function is not bounded near (0,0), hence it cannot be continuous at (0,0), hence it cannot be differentiable at (0,0). $\endgroup$ – Did Nov 29 '14 at 0:17
  • $\begingroup$ Are you sure? I read the condition $|(x,y)| < \pi$ to mean $\sqrt{x^2 + y^2} < \pi$. Near the origin, writing $r = \sqrt{x^2+y^2}$, $f$ as a function of $r$ is $r^2/\sin r$ which behaves nicely. Am I missing something? $\endgroup$ – Simon S Nov 29 '14 at 0:19
  • $\begingroup$ I guess one could interpret $\sin((x^2+y^2)^.5)$ as $\sin((x^2+y^2)^5)$... $\endgroup$ – user160738 Nov 29 '14 at 0:22
  • $\begingroup$ I did something similar to Simon S $\endgroup$ – user153009 Nov 29 '14 at 0:30
  • $\begingroup$ The partials don't exist at the origin. $\endgroup$ – Git Gud Nov 29 '14 at 0:38
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Calculating the partial derivatives at $(0,0)$ you get a limit of the form:

$$\lim_{ h\rightarrow0}\displaystyle\frac{h}{\sin |h|}$$

which does not exist.

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