Why Drop $\forall x \in \mathbb{X}$ in delta-epsilon definition at infinity? For a function $$f:X \rightarrow \mathbb{R}$$
My course notes says that:
$$\lim_{x \to \infty} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists S \in \mathbb{R}, x> S \implies |f(x)-l| < \epsilon)$$
I don't understand why we have dropped the $$\forall x \in X$$
In other words, why is it not this:
$$\lim_{x \to \infty} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists S \in \mathbb{R}, \forall x \in X, x> S \implies |f(x)-l| < \epsilon)$$
Wouldn't we want the implication to hold true for all $x$ larger than $S$, analogous to when we're dealing with the definition of limits as $x$ approaches $a$?
$$\lim_{x \to a} f(x) = l \Leftrightarrow (\forall \epsilon > 0, \exists \delta > 0, \forall x \in X, 0 < |x-a| < \delta \implies |f(x)-l| < \epsilon)$$
Otherwise, I could just find a really small $S$, smaller than a $x_1$ where $f(x_1)=l$ and if it works for one particular $x$, I could claim the limit as $x \rightarrow \infty$ is $l$, which is clearly not the intended definition of a limit as $x$ approaches infinity.
 A: I will not enter into the details of first order logic, but strictly speaking, if a variable doesn't have a quantifier ($\exists$ or $\forall$) then it is ranging over all the elements of the set.  So, again, strictly speaking both propositions are the same. In fact, strictly speaking, you can safely drop every $\forall x \in X$ whenever you know $X$ is the domain of discourse, it will just annoy your fellow mathematicians but it is right.
To see why this is the case, see this example (but remember that this is a formalism about mathematical language, so it is arbitrary in the sense that it was decided by convention to be so): if I am talking about the natural numbers and I say $x$ is even implies $x$ can be divided by $2$, one would reasonable understand that I meant, for all natural numbers $x$, $x$ is even implies it can be divided by 2. Clearly I wasn't talking about one specific $x$.
Again, just remember that this is a (intuitive) convention, nothing more than that.
A: The meaning of $x>S\implies |f(x)-l|<\epsilon$ is that "if $x>S$ then $|f(x)-l|<\epsilon$". So yes, it exactly says that for all $x$ which satisfy $x>S$ we have $|f(x)-l|<\epsilon$. An equivalent way to write the same thing:
$\forall (x: x>S)[|f(x)-l|<\epsilon]$ 
Note that in this equivalent way I didn't use $\implies$ at all. 
