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I struggling with some problems. Thank you for any help:

This function is given : $ f(x,y)=(e^x-1)\frac y{(x^2+y^2)^\alpha}\;$ , and they ask the values of $\;\alpha\;$ for which f is can be defined in origin and is differentiable at $\;(0,0)\;$ .

I tried in (1) to check when is $\;f\;$ continuous at $\;(0,0)\;$, and so the limit must exist, so if we take for example $\;y=x\;$ and let $\;x\to 0\;$ we have

$\;(e^x-1)\frac x{2^\alpha x^{2\alpha}}\to0\;$

if $\;1-2\alpha\ge 0\implies \alpha\le\frac12\;$

and we can put $\;f(0,0)=0\;$ . But I know there can be functions continuous but not differentiable so I'm stucked because the condition above is perhaps not enough.

  • I think that $\alpha \in \mathbb{R}$, so examine the cases $\alpha = 0$, $\left|\alpha\right| \leq \frac{1}{2}$, and $\left|\alpha\right| > \frac{1}{2}$. For each case try to show that $f$ can be defined in the origin so that $f$ is differentiable at $(0,0)$. – monoid Sep 21 '14 at 16:59
  • The mean value theorem tells you that $$e^x-e^0=e^\xi(x-0)$$ for some $\xi$ between $0$ and $x$. This gives you an upper bound $e|xy|$ for the numerator when $|x|\le 1$. This allows you to prove continuity (as a first goal on the path to differentiability). Note that if you go polar, you get $|xy|=r^2|\sin\phi\cos\phi|\le r^2$. – Jyrki Lahtonen Sep 23 '14 at 11:10
  • I guess what I'm saying is that $f$ is continuous at the origin, when $\alpha<1$. – Jyrki Lahtonen Sep 23 '14 at 11:15
  • Thank you very much, dear @JyrkiLahtonen. If I understood your hints then we can put $$|f(x,y)|=\left|e^\xi xy\frac1{(x^2+y^2)^\alpha}\right|\stackrel{|x|<1}\le\left|\frac{xy}{(x^2+y^2)^{ \alpha}}\right|\stackrel{\text{polar coor.}}\to$$$$\to r^2\left|\frac{\cos\theta\sin\theta}{r^{2\alpha}}\right|\le r^{2(1-\alpha)}\xrightarrow[r\to 0]{}0;;\text{, if};;1-\alpha>0\iff\alpha<1$$ Now, I can prove that when $;\alpha=1;$ the limit when $;(x,y,)\to (0,0);$ doesn't exist, but I feel this isn't enough to conclude the function is diff. only for $;\alpha <1;$ .... –  Sep 24 '14 at 21:40

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