About the additive property of little-o(h) So I encountered this definition in Salas Hille Etgen's One and Several Variables Calculus:
(This definition is for single variable case)
Definition: Let $g:\Bbb R \to \Bbb R$ be a function defined at least in some neighbourhood of $0$. We say that $g(h)$ is $o(h)$ and write $g(h)=o(h)$ to indicate that $\lim_{h\to 0}\frac{g(h)}{h}=0$.
So, after this definition, our class soon introduced a theorem:

Theorem: 
  The following are equivalent:
$1. f(x+h)-f(x)-f'(x)h=o(h)$
$2. f(x+h)-f(x)=f'(x)h+o(h)$

And my professor gave us such proof:

$f(x+h)-f(x)=[f(x+h)-f(x)-f'(x)h]+f'(x)h=o(h)+f'(x)h$

So I began to doubt: What makes it legal to operate $o(h)$ as a normal function? Although it seems so natural, we still added something wasn't in the original definition. It is not an equation, it is a sentence, isn't it?
Since the original definition write $g(h)=o(h)$ for merely implying $\lim_{h\to 0}\frac{g(h)}{h}=0$, is it okay to treat $o(h)$ like another number or function(e.g. replace $g(h)$ with $o(h)$ in the proof)?
Just like the definition of the notation of limit, it doesn't imply that it is a number. It is a sentence(description) replaced by abbreviations through the definition. 
Hence, my question are summed up in 2:


*

*Is it legal to operate $o(h)$ in such manner?

*If not, is there another way to go through this proof?


Thanks in advance.
 A: I know that writing $g(h)=\mathrm o(h)$ is a thing, but I think it is one of the most misleading notations ever $-$ when used without further explanation. So I will try to give the explanations: define $\mathrm o(h)$ as a set of functions:
$$\mathrm o(h):=\left\{f:\Bbb R\to\Bbb R \mid \lim_{h\to 0} \frac{f(h)}h = 0\right\},$$
and then $g\in \mathrm o(h)$ is what formerly was denoted by $g(h)=\mathrm o(h)$. Now we add a notational convention: any expression that contains $\mathrm o(h)$ as if it where a function is to be interpreted as "there is a function $g\in\mathrm o(h)$ replacing this occurence of $\mathrm o(h)$". So e.g. $f(x+h)-f(x)-f'(x) = \mathrm o(h)$ means
$$\forall x:\exists g_x\in\mathrm o(h):f(x+h)-f(x)-f'(x)=g_x(h).$$
Note that this might be a different $g_x$ for every $x$ (the derivative is defined point-wise). Now it might be clear why we can handle this "function" $\mathrm o(h)$ as if it where a real function. The right side above is obviously equivalent to
$$\forall x:\exists g_x\in\mathrm o(h): f(x+h)-f(x)=f'(x)+g_x(h).$$
And by our notational convention, we can write this as $f(x+h)-f(x)=f'(x)+\mathrm o(h)$.
A: I believe this is your definition (assuming, some definition of $o(\cdot)$)
$$ a = b + o(c) \text{ iff there exists }  d \text{ such that } d=o(c), \text{ and } a-b=d$$
So the proof- let $x$ be fixed. Then note that we can always define $R(h)$ to force the following two equalities at the same time,
$$f(x+h) - f(x)  - f'(x) h  = R(h) \iff f(x+h) - f(x)    =  f'(x) h+  R(h)$$
Then statement 1. makes the claim that in the LHS equation, $R(h) = o(h)$. The statement 2. (assuming you meant f'(x)h and not f'(x)) asserts again that $R(h) = o(h)$ but starting from the RHS. That these are equivalent is the above $\iff$ arrow.
