Substitution in finding limits I'm really obsessed with using substitution in order to finding limits . When we can use that ? (What's the conditions for using substitution ?) 
What's the rigorous proof which enables us to do that ? For example is that true if we say : $\lim_{x \to \infty } x - \lfloor x \rfloor $ doesn't exist so we conclude $\lim_{x \to 0 } 1/x - \lfloor 1/x \rfloor $ also doesn't exist ?
 A: A standard theorem about limits, which you can probably find a proof of in a standard calculus text is: If $\lim_{x\to a}f(x)=b$ and $\lim_{x\to b}g(x)=c$ then 
$\lim_{x\to a}g(f(x))=c$.
Proof: Let $\epsilon>0$. Since $\lim_{x\to a}f(x)=b$ and $\lim_{x\to b}g(x)=c$, there are $\delta_1,\delta_2>0$ such that $|x-a|<\delta_1$ implies $|f(x)-g|<\delta_2$, and $|x-b|<\delta_2$ implies $|g(x)-c|<\epsilon$.
So, if $|x-a|<\delta_1$, the fact that $|f(x)-b|<\delta_2|$ implies $|g(f(x)-c|<\epsilon$.  Therefore,
$$\lim_{x\to a}g(f(x))=c.$$
A: Let's discuss the case you mention: Suppose $f:(a,\infty)\to \mathbb R$ for some $a>0.$ Here is a basic substitution result for this case:
$$\tag 1\lim_{x\to \infty}f(x) = L\, \iff\, \lim_{x\to 0^+}f(1/x) = L.$$
Your question was: "is that true if we say : $\lim\limits_{x \to \infty } x - \lfloor x \rfloor $ doesn't exist so we conclude $\lim\limits_{x \to 0^+ } 1/x - \lfloor 1/x \rfloor $ also doesn't exist?" The answer is yes, by the result $(1).$ Let's go through the logic: Define $f(x) = 1/x - \lfloor 1/x \rfloor.$ Suppose $\lim\limits_{x\to \infty}f(x)$ fails to exist. Could $\lim\limits_{x\to 0^+}f(1/x)$ exist? No! If it did, then then $(1)$ would imply $\lim\limits_{x\to \infty}f(x)$ exists, contradiction. That's what the $\iff$ is all about.
So in fact $(1)$ tells us that $\lim\limits_{x\to \infty}f(x)$ fails to exist $\iff \lim\limits_{x\to 0^+}f(1/x)$ fails to exist.
Added later: General result on limits and substitution The following result covers many, but not all, cases of interest. (Missing is the case of one-sided limits.)
Thm: Let $I,J$ be open intervals. Let $s: I \to J$ be a a continuous bijection. Let $a\in I$ and set $b=s(a).$ Suppose $f: J\setminus\{b\} \to \mathbb R.$ Then $\lim_{y\to b} f(y) = L$ iff $\lim_{x\to a} f(s(x))=L.$
Notes: 1. $y=s(x)$ of course is our substitution. 2. The hypotheses imply $s^{-1}: J \to I$ is a continuous bijection. We need this for both directions, i.e., the "iff". 3. We have taken care to not assume $f$ is defined at $b.$ Hence also $f(s(x))$ is not defined at $a.$ This is important for limits in general, for example in the famous case of $\lim_{x\to 0} (\sin x)/x.$
Proof of Thm: $\implies:$  Let $\epsilon>0.$ Then there exists $\delta > 0$ such that $(b-\delta, b+ \delta)\subset J$ and $y\in (b-\delta, b+ \delta)\setminus \{b\}$ implies $f(y) \in (L-\epsilon,L+\epsilon).$
Since $s$ is continuous, there exists $\gamma > 0$ such that $(a-\gamma, a+ \gamma)\subset I$ and $x\in (a-\gamma, a+ \gamma)\setminus \{a\}$ implies $s(x) \in (b-\delta, b+ \delta)\setminus \{b\}.$ For such $x$ we have $f(s(x)) \in (L-\epsilon,L+\epsilon).$ Thus $\lim_{x\to a} f(s(x))=L$ as desired.
$\impliedby$ Briefly: This time we assume $f\circ s,$ which is defined on $I\setminus \{a\},$ has limit $L$ at $a.$ From our work above, this implies $(f\circ s)\circ s^{-1} \to L$ at $b.$ Since $(f\circ s)\circ s^{-1}=f,$ we have $f\to L$ at $b$ as desired.
