Understanding the sequential definition of a limit, particularly as n approaches infinity I am trying to formally understand the sequential definition of a limit.
My focus, in particular, is when $x \rightarrow ∞$ and when $\lim\limits_{x \to ∞}{f(x)}=∞$.
"$\delta - \epsilon$" explanations will work as well, though I am trying to focus on the sequential aspect of it.
I borrowed the following definition from here.

Let $a∈R$, let $I$ be an open interval which contains $a$, and let $f$ be a real function defined everywhere on $I$ except possibly at $a$. Then $\lim\limits_{x \to a}{f(x)}=L$
  exists if and only if $f(x_n)→L$ as $n→∞$ for every sequence $x_n∈I$∖{$a$} which converges to $a$ as $n→∞$.

Is there a way to say this without using the "$∞$" symbol?
I interpret "$x \rightarrow a$" to mean: "Every sequence, $x_n$, in $I-${$a$} that converges to $a$, as $n$ gets arbitrarily large."
The only thing I can come up with to say is: "There exists some $N$ in the natural numbers such that: For every $x_n, n>N$, $x_n$ converges to $a$."  But i'm not sure if that's correct.
Also, I interpret "$x \rightarrow ∞$" to mean: "Every sequence, $x_n$, in $I$ that diverges". 
Next, what, exactly, does it mean when $L=∞, -∞,$ and $+∞$ respectively?  
My interpretation is just that $f(x_n)$ diverges.  Again, how would I say that more formally?
Any insights here would be greatly appreciated.
 A: A sequence of real numbers $\{x_n\}_{n=1}^\infty$ converges if there is $a \in \mathbb{R}$ satisfying: for every $\varepsilon>0$ there exists a positive integer $N$ such that $|x_n-a|<\varepsilon$ whenever $n \geq N$. In the case defined above we say that $\{x_n\}_{n=1}^\infty$ converges to $a$ and we may write $x_n \to a$ as $n \to \infty$.
It is important to notice that the positive integer $N$ mentioned in the definition above will depend both on $\varepsilon>0$ and the sequence $\{x_n\}_{n=1}^\infty$. In example; The sequences $\big\{\frac{1}{n}\big\}_{n=1}^\infty$ and $\big\{-\frac{1}{3n}\big\}_{n=1}^\infty$ both converge to $0$. For each sequence, can you find positive integers $N_1$ and $N_2$ such that the definition of convergence is satisfied given $\varepsilon_1=\frac{1}{10}$ and $\varepsilon_2=\frac{1}{100}$?

A function $f$ becomes infinite as $x \to a$ if for every $M>0$ there exists $\delta>0$ such that $|f(x)|>M$ whenever $0<|x-a|<\delta$. For this limit we may write $f(x) \to \infty$ as $x \to a$.
A: Just a quick formal focus on one of the OP's questions:
Definition: A sequence $(x_n)_{n \in \mathbb N}$ of numbers in $\mathbb R$ is said to approach $+∞$ if for every $B \in \mathbb R$ the set $\{m \in \mathbb N \; | \; x_m \le B \}$ is finite. 
We also say that the limit of such a sequence is $+∞$ and write $\lim\limits_{n \to +∞} x_n = +∞$ to express this.
Definition: Let $I$ be an open interval of the form $(s,+∞) \subset \mathbb R$ and let $f$ be a real function defined everywhere on $I$. Then we say that $f$ approaches  $+∞$ as $x$ approaches  $+∞$, writing
$\tag 1 \lim\limits_{x \to ∞}{f(x)}= +∞$
to mean that for every sequence  $(x_n)_{n \in \mathbb N}$ that approaches $+∞$, if the terms $x_n$ all belong to the set $(s,+∞)$, then the corresponding sequence $(f(x_n))_{n \in \mathbb N}$ also approaches $+∞$.
Proposition: The statement $\lim\limits_{x \to ∞}{f(x)}= +∞$ is equivalent to saying that for every $B \in \mathbb R$ there exist $k \in \mathbb R$ with $(k,+∞)$ in the domain of $f$ and the image $f[\,(k,+∞)\,]$ contained in the interval $(B,+∞)$.
Calculus students certainly know, intuitively, how to graph many functions 'out at $+∞$', but these are some formal definitions.
A: $a_n$ -> a as n -> oo is defined as for all open U.
if a in U, then $(a_n)$ is eventually in U.  
$(a_n)$ is eventually in U when there is some
integer n with for all j > n, $a_j$ is in U.  
