Derivative of $x|x|$ I am trying to find the derivative of $f(x)=x|x|$ using the defition of derivative. For $x > 0$ I found that $f'(x)=2x$ and for $x<0$ the derivative is $f'(x)=-2x$. Everything is fine up to here. Now I want to check what happens when at $x=0$.
By the way, I know that $|x|$ is not differentiable at $x=0$.
So I am checking the left & right limits of $f$ when $x$ approaches $0$.


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*$\lim_{x \to 0^-}\cfrac{x|x|}{x} = \lim_{x \to 0^-}\cfrac{x(-x)}{x}=\lim_{x \to 0^-}\cfrac{(-x)}{1} = -0? = 0. $

*$\lim_{x \to 0^+}\cfrac{x|x|}{x} = \lim_{x \to 0^4}\cfrac{x(x)}{x}=\lim_{x \to 0^+}\cfrac{(x)}{1} = 0. $
I think that $f$ is not differentiable at $x=0$ since $|x|$ is not differentiable at that point. So , what do I do wrong?
Should I write something like $\lim_{x \to 0^-}\cfrac{x|x|}{x} = -0^{-}$ and $\lim_{x \to 0^+}\cfrac{x|x|}{x} =0^{+}$ so that $f'$ does not exist at $x=0$?

For information, you have the following theorem which you can use

Let $I$ be an interval, $a\in I$. If a function $f$ is continuous on $I$, has a derivative on  $I\smallsetminus\{a\}$ and if $f'(x)$ has a limit at $a$, then $f $ has a derivative at $a$, and
  $$f'(a)=\lim_{x\to a}f'(x).$$

