\begin{equation} f(x,y)=\frac{x^2\sin{y^2}}{x^2+y^4} \text{ if }(x,y) \neq (0,0) \text{ and } f(0,0)=0 \end{equation}

Prove that $f$ is differentiable at $(0,0)$.

So I started out with the definition. We have to show that there exists a function $L:\mathbb{R}^2 \to \mathbb{R}$, so that

\begin{equation} \lim_{(h,k) \to (0,0)} \frac{f(0+h,0+k)-f(0,0)-L(h)}{||(h,k)||}=\lim_{(h,k) \to (0,0)} \frac{h^2 \sin{k^2-L(h)\sqrt{h^2+k^2}}}{(h^2+k^4)\sqrt{h^2+k^2}}=...=0 \end{equation}

The problem is I don't know how to fill in the dots. I have never dealt with such a limit. I thought I could maybe use the squeeze theorem, but my mind is so foggy at the moment I don't even know what functions I should use in that case.

  • $\begingroup$ This is a suggestion I haven't thought much about, but have you tried converting it into a single variable polar limit? (I.e. $r^2=h^2+k^2$)? $\endgroup$ – KR136 Apr 21 '16 at 21:19

First note that $$||(h,k)||= \sqrt{h^2+k^2}$$

You can check if the zero operator is a viable choice for the derivative $Df(0,0)(\cdot)$ at $(0,0),$ by setting $L = 0$ and checking if the limit satisfies

$$\lim_{(h,k) \to (0,0)} \frac{f(0+h,0+k)-f(0,0)-L((h,k))}{||(h,k)||}=\lim_{(h,k) \to (0,0)} \frac{h^2 \sin{k^2}}{(h^2+k^4)\sqrt{h^2+k^2}}=0. $$

Note that

$$0 \leqslant \frac{h^2 \sin{k^2}}{(h^2+k^4)\sqrt{h^2+k^2}} \leqslant \frac{\sin{k^2}}{\sqrt{h^2+k^2}}\leqslant \frac{\sin{k^2}}{k}=k \frac{\sin k^2}{k^2}$$

By the squeeze theorem, the limit is $0$ and the operator $Df(0,0)(\cdot) = L = 0$.

  • $\begingroup$ you're a hero! Just one question, is checking if the zero operator is a viable choice an 'educated guess' or is it something you can see instantly? $\endgroup$ – Cordoba Apr 21 '16 at 21:57
  • $\begingroup$ I had a suspicion and I made estimates as I normally would to find the limit without the $L$ term. You cannot find partial derivatives at $(x,y)$ away from $(0,0)$ and take limits as $(x,y) \to (0,0)$ because they are not continuous in this case. $\endgroup$ – RRL Apr 21 '16 at 22:19

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