I need to find when $f(x)=\sqrt{|x|}$ is differentiable and find the derivative.

I found that it's differentiable when $x \neq 0$ and $f'(x)=\frac{1}{2 \sqrt{x}} \ for \ x>0$ and $f'(x)=-\frac{1}{2 \sqrt{-x}} \ for \ x<0$.

The problem is when checking the limits in 0. It is very clear to me that $\lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{x}=\infty$, but when I want to find the limit as x goes to $0^-$, I get the expression $lim_{x \rightarrow 0^-} \frac{\sqrt{-x}}{x}$ which is going to $-\infty$, but I don't know how to show it.

Please help, thank you!

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    $\begingroup$ That the derivative from the right at $x=0$ doesn't exist is enough to show that the derivative at $x=0$ doesn't exist (note $\sqrt x /x=1/\sqrt x$ for $x\ne0$). $\endgroup$ – David Mitra Dec 26 '13 at 18:02
  • $\begingroup$ How can you wite $\sqrt{-x}$ $\endgroup$ – user2369284 Dec 26 '13 at 18:06
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    $\begingroup$ user2369284, you can write $\sqrt{-x}$ when x<0 - it is not the problem. David - Thank you, understood. Still, I want to understand how to find the limit in $0^-$. $\endgroup$ – Galc127 Dec 26 '13 at 18:09
  • $\begingroup$ Notice however that a function might be differentiable in a point even if the limit of the derivative does not exist. $\endgroup$ – Emanuele Paolini Dec 26 '13 at 18:10
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    $\begingroup$ @GinKin just separate it to cases - one for x>0 and one for x<0. Then you only need to check the problematic point x=0. $\endgroup$ – Galc127 Dec 28 '13 at 15:45

By replacing $x$ with $-x$, we have

\begin{align*} \lim_{x \to 0^-} \frac{\sqrt{-x}}{x} &= \lim_{x \to 0^+} \frac{\sqrt{x}}{-x} \\ &= - \lim_{x \to 0^+} \frac{\sqrt{x}}{x} \\ &= -\infty \end{align*}

based on the computation you've already done.

  • $\begingroup$ Your answer seems to be very useful. Unfortunately, I don't understand your first step - How can I make the replacement of x and why $x \rightarrow 0^-$ changed to $x \rightarrow 0^+$? $\endgroup$ – Galc127 Dec 26 '13 at 18:11
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    $\begingroup$ @Galc127 If $x$ approaches $0$ from the left, then $-x$ approaches $0$ from the right. $\endgroup$ – user61527 Dec 26 '13 at 18:12
  • $\begingroup$ I understand that, but you didn't write -x, you wrote x and that's what confuses me. $\endgroup$ – Galc127 Dec 26 '13 at 18:13
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    $\begingroup$ @Galc127 Set $t = -x$ instead, and rewrite the other steps. $\endgroup$ – user61527 Dec 26 '13 at 18:15
  • $\begingroup$ Isn't this division by zero ? Also, why does it go to minus infinity ? $\endgroup$ – GinKin Dec 28 '13 at 11:45

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