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I'm interested in checking if the following function $$ f(x,y) = \begin{cases} \dfrac{\sin(x+y)}{\sqrt{x^2+y^2}},& \text{if } (x,y) \neq (0,0),\\ 0, &\text{if } (x,y) = (0,0). \end{cases} $$ is differentiable at the origin, and also I would like to compute its partial derivatives at the origin, if possible. For the differentiability, my claim is that $f$ is not differentiable at $(0,0)$ since it's not continuous at $(0,0)$. Indeed, if we compute the limit in polar coordinates, we obtain that $$ \lim_{r \to 0} \dfrac{\sin(r(\cos(\theta)+ \sin(\theta)))}{r} = \begin{cases} \cos(\theta) + \sin(\theta),& \text{if } \cos(\theta) + \sin(\theta) \neq 0,\\ 0, &\text{otherwise}. \end{cases} $$ therefore the limit $$ \lim_{(x,y) \to (0,0)} \dfrac{\sin(x+y)}{\sqrt{x^2 + y^2}} $$ doesn't exist. For the partial derivatives at the origin, I obtain the following $$ \dfrac{\partial f}{\partial x}(0,0) = \lim_{h \to 0} \dfrac{\sin(h)}{|h|h} = +\infty $$ and the same for the partial derivative with respect to $y$ at $(0,0)$. However, Wolfram Alpha says that the partial derivatives at the origin are both equal to $0$.

QUESTION: Is my reasoning right or have I done a mistake?

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  • $\begingroup$ I guess you are correct, but wolfram-alpha would be correct if the piecewise-function was defined to be $1$ at the origin instead of $0$. The value of $1$ is a more natural value for the function in a sense, owing to the fact that the function which is $\sin(x)/x$ for nonzero $x$ and $1$ for $x=0$ is a smooth function. $\endgroup$
    – Mike F
    Jan 17 at 18:24
  • $\begingroup$ If $f(0,0) = 1$, the partial derivatives at the origin are equal to $-\infty$, aren't they? $\endgroup$
    – Smm
    Jan 17 at 18:57
  • $\begingroup$ I don't think so. It would become a question about the derivative of $$g(x) = \begin{cases} \frac{\sin(x)}{|x|} & x \neq 0 \\ 1 & x=0 \end{cases}$$ at $x=0$. If I'm not mistaken, the right derivative would be $0$ and the left derivative would be $+\infty$. $\endgroup$
    – Mike F
    Jan 17 at 19:06
  • $\begingroup$ Anyway, I'm worried about the computation of the partial derivatives in the original problem. I think that clearly I'm right, but I would like to make sure. $\endgroup$
    – Smm
    Jan 17 at 19:11
  • $\begingroup$ I agree with your finding that, in the case of the original problem, both partials are equal to $+\infty$ at the origin. $\endgroup$
    – Mike F
    Jan 17 at 23:17
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The limit doesn't exist. Hence there can be no derivative. For let $x=0.$ Then you have $$\frac{\sin y}{|y|},$$ with limit $\pm 1$ according as $y>0$ or $<0$ as $y\to 0.$

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  • $\begingroup$ And the partial derivatives at the origin don’t exist either? $\endgroup$
    – Smm
    Jan 17 at 19:28
  • $\begingroup$ @Smm No, they technically don't, although that singularity could be easily removed. $\endgroup$
    – Allawonder
    Jan 17 at 19:39
  • $\begingroup$ The OP is still correct to say that both partials of the given function equal $+\infty$ at the origin, though. So in that sense there is a derivative. $\endgroup$
    – Mike F
    Jan 17 at 20:21

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