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

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

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    $\begingroup$ Did you try using the limit? Where are you stuck? $\endgroup$ – Arthur Oct 10 '17 at 6:17
  • $\begingroup$ Why don't you investigate the limit? Hint: put $x=0$ in the formula you gave. $\endgroup$ – Lord Shark the Unknown Oct 10 '17 at 6:17
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$\frac{f(0+h)-f(0)}{h}=h \sin(1/h)$, 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}$ ?

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Hint: $$x^2\sin(1/x)=x\times\frac{\sin(\frac{1}{x})}{(\frac{1}{x})}$$

Now try using limit $x \to 0$

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  • $\begingroup$ 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. $\endgroup$ – Arthur Oct 10 '17 at 6:22

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