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How do I check the differentiability of a $f(x) = x|x|$ at $x_0=0$?

I used the definition of a derivative and came up with \begin{align} f^\prime &= \lim_{x\to0}\dfrac{f(x)-f(x)}{x-0}\\ &= \lim_{x\to0}\dfrac{x|x|-0|0|}{x-0}\\ &= \lim_{x\to0}\dfrac{x|x|}{x}\\ &= \lim_{x\to0}|x| \end{align}

I know $f(x)=|x|$ is not differentiable at $0$. However, would this be? If one plugs in $0$ to $|x|$, one would get $0$, which is finite. But, if one takes the the limit from $-\infty$ and $+\infty$, the limits wouldn't exist.

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    $\begingroup$ Why would you take the limit to $\;\pm\infty\;$ at all? The limit is when $\;x\to0\;$ ...and that's all! $\endgroup$ – DonAntonio Nov 17 '18 at 21:28
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Why are you mentioning $\pm\infty$? What you did is correct. Since $\lim_{x\to0}\lvert x\rvert=0$, $f'(0)=0$.

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  • $\begingroup$ Would $$\lim_{x\to0^+} |x|=\lim_{x\to0^-} |x|?$$ That is why I mentioned coming from either infinity to 0. $\endgroup$ – kaisa Nov 17 '18 at 21:29
  • $\begingroup$ Both of those limits are equal to $0$. In other words, $\lim_{x\to0}\lvert x\rvert=0$. $\endgroup$ – José Carlos Santos Nov 17 '18 at 21:32

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