When I can divide and multiply in a limit by $x$ I have formulated an old question but in that case I have committed many errors in limit calculus and  I understand my doubt was not absolutely clear so I have decided to remove it (it had not answer) and now I try to formulate it better.
If I have a limit as $x$ goes to $\infty$ or $x\to 0$ I can multiply and divide by $x$ in order for instance to regain a notable limit. For instance $$\lim_{x\to 0}\frac{\sin^2{x}}{x}=\lim_{x\to 0}x\frac{\sin^2{x}}{x^2}=0$$
Now:
if I have for istance a function $f(x)$ s.t $\lim_{x\to \infty}\frac{f(x)}{x}=1$ and I have to compute $\lim_{x\to \infty}f(x)-x$, if I apply the previous argument ONLY on the term $f(x)$ I will obtain:
$\lim_{x\to \infty}x-x=0$.
$\textbf{My doubt is: }$is it allowed what i have done in the last?
 A: Others can cover counterexamples, but here I'm focusing on the computational mechanism.
Here's what you need: assuming $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exist (and in the case of division, assuming $\lim_{x \to a}g(x) \neq 0$) and are finite:
$$\lim_{x \to a}\left[f(x) \mathrel{\substack{+ \\- \\\cdot \\/}} g(x) \right] = \lim_{x \to a}f(x) \mathrel{\substack{+ \\- \\\cdot \\/}} \lim_{x \to a}g(x)\text{.}$$
In the first case, you can show that $$\lim_{x \to 0}\dfrac{\sin^2(x)}{x^2} = 1$$
thus this limit exists, and we also know that $$\lim_{x \to 0}x = 0$$
thus this limit exists, hence
$$\lim_{x \to 0}x \cdot \dfrac{\sin^2(x)}{x^2} = 0 \cdot 1 = 0\text{.}$$
What needs to be emphasized here is that the limits exist and are finite for the computation to work out.
Now, suppose you have $\lim_{x \to \infty}\dfrac{f(x)}{x} = 1$ and you want to compute $\lim_{x \to \infty}[f(x) - x]$.
Let's consider your proposed idea:
$$\lim_{x \to \infty}\left[x \cdot \dfrac{f(x)}{x} - x\right]\text{.}$$
What you had implicitly done was turned the $\dfrac{f(x)}{x}$ into $1$ by way of the limit, but the problem with that is that you're basically trying to boil it down to
$$\lim_{x \to \infty}x \cdot \dfrac{f(x)}{x} = \lim_{x \to \infty}x \cdot \lim_{x \to \infty}\dfrac{f(x)}{x} = \lim_{x \to \infty}x$$
which is NOT TRUE because $\lim_{x \to \infty}x$ is NOT finite.
A: I am not completely sure what you're asking, but if you're wondering whether
$$\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1$$
implies that
$$\lim_{x\to\infty} \left[f(x)-g(x)\right]=0,$$
then the answer is no. Consider as a counterexample $f(x) = x$ and $g(x) = x+1$.
What is true is that if $\lim_{x\to\infty} f(x)$ and $\lim_{x\to\infty} g(x)$ both individually exist, then
$$\lim_{x\to\infty} f(x)g(x) = \left(\lim_{x\to\infty} f(x)\right)\left(\lim_{x\to\infty} g(x)\right),$$
and likewise for addition, subtraction, division (by non-zero), etc.
You cannot use this rule in the first example above because the individual limits do not exist ($f(x)$ and $g(x)$ diverge to infinity).
A: $x/x = 1$ so it is not a problem.  Technically $x/x$ is undefined for $x=0$ but the limit does not depend on what happens at $0$, only what happens near $0$.
As for your second example: It sounds like you are trying to argue something like this:
\begin{align*}
\lim_{x\rightarrow \infty} f(x) - x &= \lim_{x\rightarrow \infty} x \frac{f(x)}{x} - x \\&= \lim_{x\rightarrow \infty} x \frac{f(x)}{x} - \lim_{x\rightarrow \infty} x \\&= (\lim_{x\rightarrow \infty} x) (\lim_{x\rightarrow \infty} \frac{f(x)}{x})
 - \lim_{x\rightarrow \infty} x\\
&=  \lim_{x\rightarrow \infty} x 
 - \lim_{x\rightarrow \infty} x\\
&=  \lim_{x\rightarrow \infty} x 
 - x \\
&=  0.
\end{align*}
However the step $$\lim_{x\rightarrow \infty} x \frac{f(x)}{x} - x = \lim_{x\rightarrow \infty} x \frac{f(x)}{x} - \lim_{x\rightarrow \infty} x $$ is not justified. In general the rule $$\lim_{x\rightarrow a} f(x) + g(x) = \lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x)$$ is only a valid inference if both $\lim_{x\rightarrow a} f(x)$ and  $\lim_{x\rightarrow a} g(x)$ exist (and are finite).
A: Note that :
$\lim (f(x) g(x))= ( \lim f(x) ) (\lim g(x))$ is applicable when both the limits that is $\lim f(x)$ and $\lim g(x)$ exist finitely. $\tag 1$ 
Coming to your question: 
Suppose that $\frac{f(x)}{x}\to 1$ as $x\to \infty $.
$\lim f(x)=\lim (x\frac{f(x)}{x})$. Now, you are not allowed to use $(1)$ because $\lim x$ is not finite as $x\to \infty$.
Another example when $x\to 0$, $\lim(\sin x)=\lim(x\frac{\sin x}{x})=0\times 1=0 $ because both the limits that is $\lim x$ and $\lim \frac{\sin x}{x}$ exist finitely hence $(1)$ is applicable.
