Let f(x)=casesx^21x, & x0 0, & x=0cases.

Question

Let $f(x)=\begin{cases}x^2\sin\dfrac1x, & x
e0\ 0, & x=0\end{cases}$. Then at $x=0$:

Accepted Answer

Continuity of $f$: $f(0^{\pm})=\lim_{h	o0}(\pm h)^2\sin\dfrac{1}{\pm h}=0$ (bounded sine times $h^2	o0$), and $f(0)=0$. So $f$ is continuous at $0$.

Differentiability: $f'(0)=\lim_{h	o0}\dfrac{h^2\sin(1/h)-0}{h}=\lim_{h	o0}h\sin\dfrac1h=0$. So $f$ is differentiable at $0$, with $f'(0)=0$.

For $x
e0$: $f'(x)=2x\sin\dfrac1x-\cos\dfrac1x$. As $x	o0$, $\cos\dfrac1x$ oscillates without a limit, so $f'(x)$ is not continuous at $x=0$ even though $f'(0)=0$ exists.

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