Two form of derivative $ f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ 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.
 A: The formula $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ is equivalent to 
$\lim_{h \to 0} \frac{f(x+h)-f(x)-f'(x)h}{h} = 0$.
This in turn is equivalent to the function $g(h) = f(x+h)-f(x)-f'(x)h$ satisfying  $\lim_{h \to 0} \frac{g(h)}{h} = 0$.
Such a function is referred to as 'little o' or $o(h)$, and we say $g$ is 'little o' of $h$, or simply just write $o(h)$.
That is, we abuse notation and write $f(x+h)-f(x)-f'(x)h = o(h)$, which gives rise to $f(x+h) = f(x)+f'(x)h + o(h)$.
Note: This is $o(h)$, not $O(h)$. The function $f(x) = |x|$ is not differentiable at $x=0$, but we can write $f(h)=f(0)+0.h + O(h)$, but you cannot replace the $O(h)$ by $o(h)$, if you see what I mean.
A: Let $f(x)=|x|^{3/2}$. Then $f'(0)=0$. But $\dfrac{|h|^{3/2}-0}{h}$ is not $O(h)$.  
A: By the definition of differentiability 
$$f(x+h)-f(x)=f'(x)h+o(h),\,\,h\to{0}.$$
In your's second formula must be $o(1),$ not $O(h).$
A: $f'(x)=\frac{f(x+h)-f(x)}{h}+\Big(f'(x)-\frac{f(x+h)-f(x)}{h}\Big)=\frac{f(x+h)-f(x)}{h}+o(h)$, 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 $g(x)=f'(x)$.
