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Is this function differentiable at (0,0)? .

$f(x,y)=\begin{cases}\frac{x^{2}+y^{2}}{\sqrt{\sin(x^{2}+y^{2})}}, (x,y)\neq (0,0)\\0,(x,y)=(0,0)\end{cases}$

\begin{align*} \lim_{h\mapsto 0} \dfrac{f(0+h,0)-f(0,0)}{h}=& \lim_{h\mapsto 0} \dfrac{\dfrac{h^{2}}{\sqrt{\sin(h^{2})}}}{h}\\ =& \ \text{not defined} \end{align*}

I tried to use the definition to figure out the partial derivatives. However I simply could not make it through. It looks like that the partial derivatives doesn't exist at $(0,0)$!

This was in my exam at last week. However the teacher says that it is differentiable, and I disagree. So, simply, wich one is correct?

finagle29
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capella
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1 Answers1

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Your reasoning is absolutely correct, because the limit is $\pm 1$ depending on the direction in which $h$ goes to $0$, and we know that in order for a function to be differentiable in a point it must, in particular, admit partial derivatives in that point.

Edit:

Since there seems to be some confusion, let's go through this slowly. The above computation of $\frac {\partial f} {\partial x} (0,0)$ is correct. Now, $\lim \limits _{h \to 0} \frac {\frac {h^2} {\sqrt {\sin h^2}}} h = \lim \limits _{h \to 0} \frac h {\sqrt {\sin h^2}}$ and since $\lim \limits _{h \to 0} \frac {\sin h^2} {h^2}=1$, the limit becomes $\lim \limits _{h \to 0} \frac h {\sqrt {\frac {\sin h^2} {h^2}}} \frac 1 {\sqrt h^2} = \lim \limits _{h \to 0} \frac h {|h|} = \pm 1$, exactly as you say yourself.

It is ridiculous that you were right and your teacher was wrong on such a simple issue. In general, don't trust people because they represent authority, trust clear reasoning.

Alex M.
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