1
$\begingroup$

I have the following function: $f: \mathbb{R}^2 \to \mathbb{R}$ with $f(x,y) = \frac{x^2y}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0) = 0$. Now I want to proof that $f$ is differentiable in $(0,0)$, but I get stuck with an $\epsilon-\delta$ proof for the limit \begin{equation} \lim_{(k,h) \to (0,0)} \frac{f(0+k,0+h) - f(0,0) - kf_1(0,0) - hf_2(0,0)}{\sqrt{k^2 + h^2}} = 0 \end{equation} I think that $f_1(0,0) = f_2(0,0) = 0$, because $\lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{\frac{h^2 \cdot 0}{h^2 + 0}-0}{h} = 0$. And the same for $y+h$.

So the actual limit I want to solve is \begin{equation} \lim_{(k,h) \to (0,0)} \frac{\frac{k^2h}{k^2+h^2}}{\sqrt{k^2+h^2}} = \lim_{(k,h) \to (0,0)} \frac{k^2h}{(k^2 + h^2)^{\frac{3}{2}}} = 0 \end{equation} Perhaps I made a mistake earlier on, but I can't seem to give a right proof of the existence of the limit...

$\endgroup$
5
$\begingroup$

The function $f$ is not differentiable at $(0,0)$. If it was, then, since $f_x(0,0)=f_y(0,0)=0$, then $f'(0,0)$ would be the null function. In other words,$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{(x^2+y^2)^{3/2}}=0.$$But if $x=y>0$, $\frac{x^2y}{(x^2+y^2)^{3/2}}=\frac1{2\sqrt2}$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.