Property of asymptotic notation little $o$ in limits. For example, we have a limit
$$\lim_{x\to1} \:\frac{\ln(x)}{1-x}=A$$
and we know that as $x\rightarrow 0$ we have  $$\ln(x)= x-1 +o(x-1), \qquad \text{(1)}$$ which by definition means:
$$\lim_{x\to1} \:\frac{\ln(x)-(x-1)}{x-1}=0.$$
I know that by knowing this definition and using some limit properties we can get the answer $A$.
But are we allowed to just change the limit (by plugging in (1))? Namely to the following one:
$$\lim_{x\to1} \:\frac{x-1 +o(x-1)}{1-x}=-1=A$$
and proceed to solve this limit. What are the justifications by their limit definitions? 
This example might be too simple, but what I want to point out is that, most of the times it is easier to plug in (1) and solve the limit than to write out the definition of (1) and do algebraic manipulations and by using some limit properties to get the answer $A$.
 A: Let's first rewrite the definition of what it means for something to be ‘little-oh’ of something else. We say $f$ is little-oh $g$ as $x$ goes to $a$ iff $$\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = 0,$$ and we write $$f(x) = o(g(x)) \quad \text{as } x\rightarrow a.$$ Even more formally, $o(x)$ represents a set, and we should write $f(x) \in o(g(x)).$ Hence when we write \begin{equation} \log(x) = x-1 + o(x-1), \end{equation} what we really mean is \begin{equation} \log(x) = x-1 + f(x-1), \end{equation} for $f(x-1) \in o(x-1)$ as $x-1 \rightarrow 0.$ Now writing out your limit, we get $$\lim_{x\rightarrow 1} \frac{\log(x)}{1-x} = \lim_{x-1\rightarrow 0} \frac{x-1+f(x-1)}{1-x} \\
= -1 - \lim_{x-1\rightarrow 0} \frac{f(x-1)}{x-1} = -1,$$ which is the same result as if you had simply plugged in $\log(x) = x-1 + o(x-1).$
From this, you should be able to see why you are justified in using the $o(x)$ notation: because it's a useful shorthand for something more formal.  As long as you remember what the notation stands for, you should be able to manipulate it sensibly.

Edit: further clarifications.  One thing that might also help is if you consider the change of variables $x-1 \mapsto y.$  Copying the above paragraphs with this transformation gives the following.

When we write \begin{equation} \log(y+1) = y + o(y), \end{equation} what we really mean is \begin{equation} \log(y+1) = y + f(y), \end{equation} for $f(y) \in o(y)$ as $y \rightarrow 0.$ Now writing out your limit, we get $$\lim_{y+1\rightarrow 1} \frac{\log(y+1)}{-y} = \lim_{y\rightarrow 0} \frac{y+f(y)}{-y} \\
= -1 - \lim_{y\rightarrow 0} \frac{f(y)}{y} = -1,$$ which is the same result as if you had simply plugged in $\log(y+1) = y + o(y).$


Edit (ii): clarification regarding the definition of a limit.  By definition, $$\lim_{y\rightarrow b} h(y) = B \Longleftrightarrow \\ (\forall \epsilon > 0) (\exists \delta > 0)(\left\|{y-b}\right\| < \delta \Rightarrow  \left\|{h(y)-B}\right\| < \epsilon).$$
