Let's consider the following function:

$$f(x,y)=\begin{cases} (x^2+y^2)\sin\left(\dfrac{1}{x^2+y^2}\right) & \text{if }x^2+y^2\not=0 \\{}\\ 0 & \text{if }x=y=0 \end{cases}$$

I know that $f_x$ and $f_y$ are not continuous at $0$. How to prove that $f$ is differentiable at $0$?

  • $\begingroup$ Is that $\;-1\;$ on the sine's argument or on the sine itself? $\endgroup$ – Timbuc May 21 '15 at 20:39
  • $\begingroup$ @Timbuc Why is the extra space necessary?\ $\endgroup$ – Cookie May 21 '15 at 20:40
  • $\begingroup$ @dragon Doesn't it look nicer and neater? $\endgroup$ – Timbuc May 21 '15 at 20:40
  • $\begingroup$ I assume the exponent is on the argument of the sine function. Correct? $\endgroup$ – Mark Viola May 21 '15 at 20:45
  • $\begingroup$ Does the $^{-1}$ belong to sin as a whole or just the argument $\endgroup$ – grdgfgr May 21 '15 at 20:46


Change to polar coordinates and show

$$\lim_{r\to 0}\frac{r^2\sin(r^{-2})-0}{r}=0$$

Hint 2: The sine function is bounded.

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