# Proving that $f$ is differentiable at $0$

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 $0$?

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

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