# Can't understand how to find a limit $f(x)=\sqrt{|x|}$

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.

• 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 $x\ne0$). – David Mitra Dec 26 '13 at 18:02
• How can you wite $\sqrt{-x}$ – user2369284 Dec 26 '13 at 18:06
• 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 $0^-$. – Galc127 Dec 26 '13 at 18:09
• Notice however that a function might be differentiable in a point even if the limit of the derivative does not exist. – Emanuele Paolini Dec 26 '13 at 18:10
• @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. – Galc127 Dec 28 '13 at 15:45

By replacing $x$ with $-x$, we have
• 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 $x \rightarrow 0^+$? – Galc127 Dec 26 '13 at 18:11
• @Galc127 If $x$ approaches $0$ from the left, then $-x$ approaches $0$ from the right. – user61527 Dec 26 '13 at 18:12
• @Galc127 Set $t = -x$ instead, and rewrite the other steps. – user61527 Dec 26 '13 at 18:15