$∀$ quantifier in Epsilon delta definition of limits given $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$$
means by Epsilon Delta definition :
∀ $ε>0$, ∃ $δ>0$ such that  $0<|x−a|<δ$ implies $|(f(x)-L|< ε$
my question is about ∀ $ε>0$ , how we can make sure that any picked distance form L will be defined in the range of function ? I looking at ∀ $ε>0$ as "Choose any distance you want from L " and i think this is true for a function with range $\left(-\infty \:,\:\infty \:\right)$.
I mean there is no such a restriction to this in pierce definition or we just assume that $ε$ must be a very small number.

used function :
$f(x)=\left\{x>2:2+\frac{2}{\left(x-1\right)^2}\right\}$
 A: Suppose the condition holds when $\epsilon$ is small, say when $\epsilon <r$. Then, for any $\epsilon >0$ take $\epsilon_1 \in (0, \min \{r, \epsilon\})$ and find $\delta$ such that $|f(x)-L| < \epsilon_1$ for $0 <|x-a| <\delta$. This is possible because $\epsilon_1 <r$. But now we also have $|f(x)-L| < \epsilon$. So it is possible to find a $\delta$ for any $ \epsilon >0$. This has nothing to do with the range of $f$. 
A: An example picture/graph helps to 'bring to life' Paramanand Singh's comment.
When 'warming up' to an epsilon/delta, you can play around with real numbers.
Suppose $\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$
with $a = 2$ and $L = 3$. so that you are looking at
$$\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 3$$
for some function $f$.
If you are presented with an actual number $\varepsilon = .75$, your analysis shows that with $\delta =.5$,
$$ f\left(\,(\frac{3}{2},\frac{5}{2}) \setminus \{2\}\,\right) \subset (\frac{9}{4},\frac{15}{4})$$
You can graph  the corresponding $\varepsilon/\delta$ rectangle:

and you can thereby 'control the fluctuations' of $f$ by 'keeping its action inside that rectangle'.
If you are given another actual number, say $\varepsilon = .25$, you can 'accept the challenge' once again, and come up with a $\delta$ and the 'action inside rectangle'.
You now decide to accept an arbitrary symbol $\varepsilon \gt 0$, finding a $\delta$ such that
$$ f\left(\,(2-\delta,2+\delta) \setminus \{2\}\,\right) \subset (3-\varepsilon, 3+\varepsilon)$$
In general (removing the picture), $\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$ if for every $\varepsilon \gt 0$ there exist a $\delta \gt 0$ such that
$$\tag 1 f\left< \;(a-\delta,a+\delta) \setminus \{a\}\;\right> \subset (L-\varepsilon, L+\varepsilon)$$
