Understanding limits at infinity with regard to the definition of a limit This is sort of a follow up to my previous question
Say you have 
$$
\lim_{x\to +\infty} f(x)
$$
where $f : \mathbb{R} \to \mathbb{R} , x \in \mathbb{R}$
What exactly does this mean? From the last question I asked I understand $+\infty$ to be a concept (I also read this) meaning "a number that is arbitrarily large" (taken from that page).
The definition of the limit of a function, as I pointed out in my last question, specifies that the point which $x$ approaches must be a limit point; however, again from my last question, I came to understand that $+\infty$ is not a limit point (considering how I described it above this makes sense, as its not a number but a concept), so how can we take the traditional limit of this?
I have thought about this and have come to the conclusion that a traditional $\epsilon , \delta$ proof wouldn't make sense, and that this simply means "what is the value of $f(x)$ as $x$ gets really really big. Is this correct thinking?
Thanks in advanced!
 A: The statement that 
$$\lim_{x\to\infty}f(x)=L\tag{1}$$
means precisely this: 

for each $\epsilon>0$ there is a real number $x_\epsilon$ such that $|f(x)-L|<\epsilon$ whenever $x\ge x_\epsilon$.

That’s the actual definition of $(1)$. As you can see, it’s very similar to the $\epsilon$-$\delta$ definition of $$\lim_{x\to a}f(x)=L\;,$$ but with $|x-a|<\delta$ replaced by $x\ge x_\epsilon$. You don’t have an actual notion of ‘distance from $x$ to $\infty$’, but just as making $\delta$ smaller makes $|x-a|<\delta$ a stronger statement about the closeness of $x$ to $a$, so making $x_\epsilon$ larger makes $x\ge x_\epsilon$ a stronger statement about the ‘closeness’ of $x$ to $\infty$, speaking informally. (As I said in comments on the earlier question, this can be made more formal in a more general topological setting.)
A: If you got the idea that $+\infty$ is the concept of arbitrarily large numbers, then you got the wrong idea. There are relationships, though; for example, $+\infty$ represents the 'limiting' behavior of "arbitrarily large positive numbers" in the same fashion that $0$ represents the 'limiting' behavior of "arbitrarily small positive numbers".
The right (IMHO) way treat $+\infty$ in the way it's used in calculus is by the extended real numbers. The extended real numbers comprise all ordinary real numbers, and two additional numbers which we call $+\infty$ and $-\infty$.
It turns out that, in this setting, $\lim_{x \to +\infty} f(x) = L$ or $\lim_{x \to a} f(x) = +\infty$ means exactly the same thing as any other limit... once we pass to the general topological notion of limit.
In the more general setting, rather than having an $\epsilon$ or a $\delta$ that says how far some number can be from some other number, we instead consider the idea of an open set or an open neighborhood. The definition of a limit is:

$\lim_{x \to a} f(x) = L$ if and only if, for every open neighborhood $U$ containing $L$, there exists an open neighborhood $V$ containing $a$, such that for every $x \in V$ such that $x \neq a$, we have $f(x) \in U$.

In the standard real numbers, we can choose to define "open neighborhood" to mean "open interval". So every neighborhood looks like $(a,b)$: e.g. $(L - \epsilon, L + \epsilon)$ is an open neighborhood when $\epsilon > 0$. Hopefully it's clear how the general notion of limit reduces to the version you learned in elementary calculus!
When using the extended real numbers, we also take $(a, +\infty]$ and $[-\infty, a)$ as open neighborhoods (where $a$ is an ordinary real number). So,
$$\lim_{x \to +\infty} f(x) = L$$
Means, after a little bit of rewriting and simplification,

For every $\epsilon > 0$, there exists an $N$ such that for every (ordinary) real number $x > N$ we have $|f(x) - L| < \epsilon$

which is exactly what you learned in elementary calculus!
