Question: Why I can write formula derivative $$ f'(x) = \lim\_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ in this form: $$ f'(x)=\frac{f(x+h)-f(h)}{h}+O(h)$$
I know, that it's easy but unfortunately I forgot.

Answer: 
Let $f(x)=|x|^{3/2}$. Then $f'(0)=0$. But $\dfrac{|h|^{3/2}-0}{h}$ is not $O(h)$.
$f'(x)=\frac{f(x+h)-f(x)}{h}+\Big(f'(x)-\frac{f(x+h)-f(x)}{h}\Big)=\frac{f(x+h)-f(x)}{h}+o(h)$, assuming $f'(x)$ exists (since $\Big(f'(x)-\frac{f(x+h)-f(x)}{h}\Big)$ goes to $0$ as $h\to 0$). It's true that if something is $o(h)$ it is also $O(h)$, so it's also true that $f'(x)=\frac{f(x+h)-f(x)}{h}+O(h)$. Also if $g(x)=\frac{f(x+h)-f(x)}{h}+o(h)$, then by taking $h\to 0$, we get that $g(x)=f'(x)$.