Definition of Differentiability (equivalent form) 
Suppose $E$ is an open set in $\mathbb R^n$, $f$ maps $E$  into $\mathbb R^m$, and $x\in E.$   If there exists a linear transformation $A$ of  $\mathbb R^n$ into $\mathbb R^m$ such that
$$ \lim_{h\to 0} \frac{|f(x+h)-f(x)-Ah|}{|h|}=0$$
then we say that $f$ is differentiable at $x$, and we write
$$f'(x)=A$$

Question: Can we rewrite the above definition to the following form:
$$ f(x+h)-f(x)=f'(x)h + r(h)$$
where the remainder $r(h)$ satisfies
$$ \lim_{h\to 0} \frac{|r(h)|}{|h|}=0. $$ If so, how to justify this.
 A: We can. And in fact, it's close to the version I prefer, personally: $f$ is differentiable at $x\in E$ if there exist a linear map $f'(x):\mathbb R^n\to\mathbb R^m$ and a map $r:E-x\to\mathbb R^n$ such that for all $h\in\mathbb R^n$ for which $x+h\in E$ we have that
$$f(x+h)=f(x)+f'(x)h+r(h),$$
where $\lim_{h\to0}\frac{\vert r(h)\vert}{\vert h\vert}=0$ (that is, $r\in o(\vert h\vert)$). By $E-x$ I mean $E$ translated by $-x$.
This is Taylor's theorem for once differentiable functions, which is why I like this version: It says that a function is differentiable if there is a first degree Taylor "polynomial" with which  it can be approximated.
Justifying this version (or yours, which is really the same via a simple rearrangement of the equation) is as simple as solving for $r(h)=f(x+h)-f(x)-f'(x)h$ and inserting it into the limit condition placed on $r$. We obtain the usual definition of differentiability. And the other way around we just choose $r(h)=f(x+h)-f(x)-f'(x)h$ and quickly see that $f(x+h)=f(x)+f'(x)h+r(h)$, and that $r$ satisfies the limit condition by definition.
A: If we have $$\lim_{h\to 0} \frac{|f(x+h)-f(x)-Ah|}{|h|}=0$$ then our left hand side will be equal to some function $|g(h)|$ some function $r: U\to \mathbb{R}^m$ with $U$ being some small neighborhood near $0$. in otherwords, $$\frac{|f(x+h)-f(x)-Ah|}{|h|}=|g(h)|$$
Then we can multiply the $|h|$ away to get $$|f(x+h)-f(x)-Ah|=|h||g(h)|$$
If the previous equation is true then there ought to be a function $g(h)$ such that we have $$f(x+h)-f(x)-Ah=g(h)|h|$$ or $$f(x+h)-f(x)=Ah+g(h)|h|$$ By our previous supposition, we must have $\lim_{h\to 0} g(h)=0$ if we let $r(h)=g(h)|h|$ then we have $$\lim_{h\to 0} \frac{r(h)}{|h|}=\lim_{h\to 0} g(h)=0$$ meaning that if our function $f$ satisfies the first definition, then it satisfies our new one.
