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Let $f:\mathbb{R^2} \rightarrow \mathbb{R}$ be a function defined as \begin{align*} f(x,y) = \begin{cases} (x^2+y^2){\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)} &\quad\text{if } (x,y) \ne(0,0) \\ 0 &\quad\text{otherwise.} \\ \end{cases} \end{align*}

Prove that $f$ is differentiable at $(0,0)$ but its partial derivatives aren't continuous at $(0,0)$.


For the first part, I proved it by definition of differentiability at that point. So, I want that

\begin{align*} \lim_{(x,y)\to\ (0,0)} \frac{f(x,y) - f(0,0) - \frac{\partial f}{\partial x}(0,0)x - \frac{\partial f}{\partial y}(0,0)y}{\Vert (x,y) - (0,0)\Vert} = 0 \end{align*}

By definition, $f(0,0) = 0$. Then i calculate $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$:

\begin{align*} \frac{\partial f}{\partial x}(0,0) = \lim_{h\to\ 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h\to\ 0} \frac{h^2\sin\left(\frac{1}{\mid h \mid}\right)}{h} \leq\lim_{h\to\ 0} \frac{h^2}{h} = 0 \end{align*} \begin{align*} \frac{\partial f}{\partial y}(0,0) = \lim_{h\to\ 0} \frac{f(0,h) - f(0,0)}{h} = \lim_{h\to\ 0} \frac{h^2\sin\left(\frac{1}{\mid h \mid}\right)}{h} \leq\lim_{h\to\ 0} \frac{h^2}{h} = 0 \end{align*} And, \begin{align*} \lim_{(x,y)\to\ (0,0)} f(x,y) = \lim_{(x,y)\to\ (0,0)} (x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) \leq \lim_{(x,y)\to\ (0,0)} (x^2+y^2) = 0 \end{align*}

Then the limit is 0 so f is differentiable at $(0,0)$. To prove the partials aren't continuous, I calculate $\frac{\partial f}{\partial x}(x,y)$ using derivative rules around $(0,0)$:

\begin{align*} \frac{\partial f}{\partial x}(x,y) = \begin{cases} 2x\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)-x\frac{\cos\left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}} &\quad\text{if } (x,y)\ne(0,0) \\ 0 &\quad\text{otherwise.} \\ \end{cases} \end{align*}

Which i have to prove is not continuous but I can't find the right curves to accomplish this. Any suggestions?

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marked as duplicate by José Carlos Santos, user99914, Namaste calculus Mar 5 '18 at 0:27

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You can try to evaluate your partial derivative at $(\frac{1}{n\pi},0)$. The $\sin$ is then $0$. and the $\cos$ is +1 or -1. This points tend to zero but the limit of the partial derivative evaluated at this point is not $0$.

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  • $\begingroup$ Thank you very much!! :) $\endgroup$ – jscherman Mar 5 '18 at 2:38

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