Why can we consider only arbitrarily small epsilon in the definition of the limit of a sequence? In the definition of the limit of a sequence I thought that we must consider all $\epsilon > 0$, but I learned that it suffices to let $\epsilon$ be an arbitrary small positive number. That is we need only consider $\epsilon \in (0, x)$. I'm trying to found out how you can get this from the definition of the limit of a sequence.
 A: OK, this seems obvious now. From what @Did pointed out, if for any $\epsilon \in (0,x)$ we can show there exists $N_\epsilon$ such that $\forall n \ge N_\epsilon$
$$|a_n - L| < \epsilon$$
then for any $\epsilon' \ge \epsilon$ there exists $N_\epsilon$ such that $\forall n \ge N_\epsilon$
$$|a_n - L| < \epsilon \le \epsilon'$$
So we can conclude that for any $\beta > 0$ there exists a $N'$, (namely $N_\epsilon$), such that $\forall n \ge N'$
$$ |a_n - L| < \beta $$
A: From Point Set Topology, we have the following definition of a limit point of a set:
$\zeta$ is a limit point of $S \subseteq \mathbb{R}$ if every deleted neighbourhood of $\zeta$ contains at least one point of $S$. 
In symbols: 
$[N_\epsilon (\zeta)- \left\{ \zeta \right\}] \cap S \neq \phi$  $ \forall \epsilon$ $>$ $0$
Now, consider the range set of the function $f(x)$, i.e $R(f(x))$ with limit point $\zeta$ .
By now, I would like to introduce a theorem:
If $\zeta$ is a limit point of a set $S\subseteq \mathbb{R}$, then every nbd of $\zeta$ contains infinitely many points of $S$

Let us now apply the notion to the range set of $f(x)$.
No matter how small the value of $\epsilon$ is, it still contains infinitely many values of the set. Thus, we can take arbitrarily small positive value of $\epsilon$. I hope, this answer will be helpful.
