I need a solution for this question. I've been trying out this question for days and I haven't been able to find out its solution yet. And some explanation would help too.
Show that the function f defined by: $$f(x):= \begin{cases} x^2\sin(1/x) &:\text{if $x \ne 0$} \\ 0 &:\text{if $x=0$} \end{cases}$$ is differentiable at $x=0$, and that $f'(0)=0$.
Hint: We have $-x^2 \le f(x) \le x^2$, hence $$ -x \le \frac{f(x) - f(0)}{x-0} = \frac{f(x)}x \le x $$ for $x \ne 0$.
Hints:
$$\bullet\;\;\text{If $\,f,g\,$ are two functions defined in some punctured neighborhood I$_0$ of $\;x_0\;$ and s.t.:}$$
$$\begin{align*}(1)&\;\;\lim_{x\to x_0}f(x)=0\\{}\\ (2)&\,\exists\,M\in\Bbb R\;\;s.t.\;\;|g(x)|\le M\;\;\forall\,x\in\text{I}_0\end{align*}$$
$\;\;\;\;\;\;\;\;\;\;\;$then $\;\displaystyle{\lim_{x\to x_0}f(x)g(x)=0}$
$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\bullet\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lim_{x\to x_0}\frac{f(x)-f(0)}{x}=\lim_{x\to x_0}\;x\sin\frac1x\;\;\ldots\ldots$$
