Prove $f(x)$ differentiable at the origin [closed]

$$f(x) = \cases{ x^2\sin(1/x) & if x \neq 0 \\ 0 & if x = 0}$$ justify that $f(x)$ is differentiable at the origin using $f'(x)=\lim_{h\to0} \frac{f(x+h)-f(x)}h$

closed as off-topic by Hans Lundmark, Aqua, Xander Henderson, jvdhooft, DaveOct 10 '17 at 13:20

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• Did you try using the limit? Where are you stuck? – Arthur Oct 10 '17 at 6:17
• Why don't you investigate the limit? Hint: put $x=0$ in the formula you gave. – Lord Shark the Unknown Oct 10 '17 at 6:17

$\frac{f(0+h)-f(0)}{h}=h \sin(1/h)$, hence
$|\frac{f(0+h)-f(0)}{h}| \le |h|$.
What can you say about $\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}$ ?
Hint: $$x^2\sin(1/x)=x\times\frac{\sin(\frac{1}{x})}{(\frac{1}{x})}$$
Now try using limit $x \to 0$
• We are not trying to show that the limit of $f$ as $x\to 0$ is $0$, but rather that the function is differentiable at the origin. – Arthur Oct 10 '17 at 6:22