Let $f(x) := x^2 \sin \frac 1 x, f(0)=0$. Show $f$ is differentiable on $\mathbb R$. Let $f: \mathbb R \rightarrow \mathbb R$ defined by $$f(x) := \begin{cases}x^2 \sin \frac 1 x\ & x \neq 0\\ 0\  & x = 0\end{cases}$$
Show $f$ is differentiable on $\mathbb R$:
Let $\epsilon > 0$.
$x_0 \neq 0$: $\left\lvert\frac {x^2 \sin \frac 1 x - x_0^2 \sin \frac 1 {x_0}}{x-x_0} \right\rvert \le \epsilon$ 
I'm thinking that since the fraction is continuous there exists $\delta > 0$ satisfying the above for $x \in [-\delta,\delta] / \{0\}$ ?
$x_0 = 0$: $\left\rvert\frac {x^2 \sin \frac 1 x - 0} {x} \right\rvert \le \epsilon$ 
How can I evaluate this limit ? I cannot say the fraction is continuous at $0$ ?
Also I must find the derivative of $f$ and show it is not continuous at $x=0$, but I'm stuck.
 A: Proving differentiability at $0$ is equivalent to proving that $\lim \limits_{x\to 0}\left(\dfrac {f(x)-f(0)}{x-0}\right)$ exists and is finite.
This is turn is equivalent to proving the same hods for $\lim \limits_{x\to 0}\left(x\sin \left(\dfrac 1 x\right)\right)$.
The last statement above is equivalent to
$$\forall \varepsilon >0\,\exists \delta >0\,\forall x\in D_f\left(0<|x-0|<\delta \implies \left|x\sin \left(\dfrac 1 x\right)\right|<\varepsilon\right).$$
To prove it recall that $\forall \alpha \in \mathbb R(|\sin(\alpha)|<1)$. Can you find $\delta$ given $\varepsilon$?
A: Lemma: if $\;h,g\;$ are real functions defined in some neighborhood $\;D_0\;$ of $\;x_0\in\Bbb R\;$ s.t.
$$\begin{cases}\lim_{x\to x_0}h(x)=0\\{}\\\exists\,M\in \Bbb R\;\;s.t.\;\;\forall\,x\in D_0\;,\;\;|g(x)|\le M\end{cases}\;\;,\;\;\text{then}\;\;\lim_{x\to\ x_0}h(x)g(x)=0$$
The proof is boringly simple, and it gives you what you want since
$$\frac{f(x)-f(0)}{x-0}=x\sin\frac1x$$
and $\;\sin\frac1x\;$ is bounded in a neighborhood of zero....
