How to calculate $f'(0)$ from $f(x)=x|x|$ directly from the definition of derivatives as a limit?
So $\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\frac{(x+h)|x+h|-|x|x}{h}=\lim\limits_{h\to 0}\frac{x|x|+x|h|+|x|h+h|h|}{h}$
Then I fail to continue as I have no idea how to deal with the absolute signs.
Could someone help?
First note that $|x+h|\neq|x|+|h|$. This is, as a friend of mine would say, "cowboy math" (no offense to any cowboy-mathematicians out there).
Next, since you are trying to calculate the derivative at $x=0$, you can replace $x$ in your limit with $0$. This makes things a bit easier. To solve the limit, look at the one-sided limits. If they are the same, then the limit exists and is equal to both.
$$\lim_{h\to 0^+}\frac{(0+h)|0+h|-0}{h}$$
$$\lim_{h\to 0^-}\frac{(0+h)|0+h|-0}{h}$$
In the first limit, $h>0$, so you can replace $|h|$ with $h$. In the second, $h<0$, so you can replace $|h|$ with $-h$. Show that both these limits are equal.