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I have to determine differentiability at $(0,1)$ of the following function: $$f(x,y)=\frac{|x| y \sin(\frac{\pi x}{2})}{x^2+y^2}$$ The partial derivatives both have value $0$ at $(0,1),$ and both are continuous on that point (I think I've got this part right), so the function must be differentiable at $(0,1).$ But when I checked for differentiability using the definition, the limit that should be $0$ doesn't exist, so I assume I'm doing something wrong when computing the limit. The following limit has to be $0$ if the function is differentiable at that point $$\lim_{x,y\to(0,1)} \frac{|f(x,y)|}{\|(x,y)-(0,1)\|}$$

Doing the change $w=y-1$ we have: $$\lim_{x,y\to(0,0)} \frac{|x (w+1)\sin(\frac{\pi x}{2})|}{(x^2+(w+1)^2)\sqrt{x^2+w^2}}$$ and then computing the limit along the line $x=w,$ it has the value $\pi /2\sqrt{2}$, which contradicts that the limit is $0.$

What am I doing wrong?

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We have that

$$\frac{|x (w+1)\sin(\frac{\pi x}{2})|}{(x^2+(w+1)^2)\sqrt{x^2+w^2}}=\frac \pi 2\left|\frac{\sin(\frac{\pi x}{2})}{\frac{\pi x}{2}}\right|\frac{| w+1)|}{(x^2+(w+1)^2)}\frac{|x^2|}{\sqrt{x^2+w^2}} \to \frac \pi 2\cdot 1\cdot 1\cdot 0=0$$

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I don't see the problem. On the line $x=w,$ the expression is

$$\frac{O(|x|^2)}{(2x^2)^{1/2}}=\frac{O(|x|^2)}{\sqrt 2 |x|}\to 0.$$

I think $Df(0,1)$ is indeed the $0$ transformation.

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  • $\begingroup$ Where did that expression come from? To compute the limit along x=w I substitute w for x and then do the limit as x tends to 0. $\endgroup$ – Nonimous Dec 3 '19 at 17:29
  • $\begingroup$ Sorry, I left $w$ in the exression by mistake. I edited it. $\endgroup$ – zhw. Dec 3 '19 at 17:36
  • $\begingroup$ I see my mistake now, quite a stupid one, thank you for the response. $\endgroup$ – Nonimous Dec 3 '19 at 19:11

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