How does one extend the limit concept to extended real numbers? Consider a function $f: \mathbb{R}\to [-\infty, + \infty]$. Is it possible to define
$$\lim_{x \to a} f(x)$$
where $a \in \mathbb{R}$ without appealing to a metric on $[-\infty, + \infty]$?
If yes, how?
 A: Eric Yau's comment goes straight to the point: the extended real line $\overline{\mathbb{R}} = [-\infty, +\infty]$ is a topological space, and it is common knowledge how to extend the definition of $\lim_{x\to a}f(x) = b$ for functions $f \colon X \to Y$ between metric spaces to the case where $Y$ (and possibly also $X$) is a topological space on which a metric may not have been given.
An answer to the question can be assembled from the Wikipedia page Limit of a function [which incidentally needs some tidying up]:

Functions on topological spaces
Suppose $X,Y$ are topological spaces with $Y$ a Hausdorff space. Let $p$ be a limit point of $\Omega \subseteq X$, and $L \in Y$. For a function $f \colon \Omega \to Y$, it is said that the limit of $f$ as $x$ approaches $p$ is $L$ (i.e., $f(x) \to L$ as $x \to p$) and written
  $$
\lim_{x \to p}f(x) = L 
$$
  if the following property holds:
  
  
*
  
*For every open neighborhood $V$ of $L$, there exists an open neighborhood $U$ of $p$ such that $f(U \cap \Omega − {p}) \subseteq V$.
  
  
  This last part of the definition can also be phrased "there exists an open punctured neighbourhood $U$ of $p$ such that $f(U \cap \Omega) \subseteq V$". 
Limits involving infinity
[...]
  
  
*
  
*a neighborhood of $-\infty$ is defined to contain an interval $[-\infty,c)$ for some $c\in\mathbb{R}$,
  
*a neighborhood of $\infty$ is defined to contain an interval $(c,\infty]$ where
  $c\in\mathbb{R}$, and
  
*a neighborhood of $a\in\mathbb{R}$ is defined in the normal way [for the] metric space $\mathbb{R}$.
  
  
  [...] $\overline{\mathbb{R}}$ is a topological space and any function of the form
  $f\colon X\to Y$ with $X,Y\subseteq\overline{\mathbb{R}}$ is subject to the topological
  definition of a limit. Note that with this topological definition, it is easy to define infinite limits at finite points, which have not been defined above in the metric sense.

I said that this is "common knowledge", but it seems remarkably hard to find a clear statement of it in print! (One could always print out the Wikipedia page, I suppose!) :)
As indicated in my
comment, I thought that the definition of the
limit of a function between general topological spaces
might be buried somewhere in a discussion of filter convergence (or net
convergence); but I couldn't find it there, either (and I looked in
quite a number of books). I hope that someone can supply the lack of an
authoritative reference.

A clear general definition of the meaning of
$\lim_{x\to x_0} f(x) = y$ can be given by condensing extracts from
pages 48--62 of Horst Schubert, Topology (Macdonald 1968):

A non-empty system $\mathscr{B}$ of subsets of the set $X$ is called
  a filter basis (on $X$) if the following conditions are satisfied:
FB 1 The intersection of two sets from $\mathscr{B}$
  contains a set from $\mathscr{B}$.
FB 2 The empty set does not belong to $\mathscr{B}$.
A non-empty system $\mathscr{F}$ of subsets of the set $X$ is called
  a filter (on $X$) if the following conditions are satisfied:
F 1 Every superset of a set from $\mathscr{F}$ belongs to $\mathscr{F}$.
F 2 The intersection of two sets from $\mathscr{F}$ belongs to $\mathscr{F}$.
F 3 The empty set does not belong to $\mathscr{F}$.
The neighbourhoods of a point $x$ form a filter, the
  neighbourhood filter of $x$. Every filter basis $\mathscr{B}$
  on $X$ determines a filter $\mathscr{F}$ on $X$.
Let $\mathscr{F}$ and $\mathscr{G}$ be two filters on the set $X$.
  If every set of $\mathscr{F}$ belongs to $\mathscr{G}$, then
  $\mathscr{F}$ is said to be coarser than $\mathscr{G}$ and
  $\mathscr{G}$ finer than $\mathscr{F}$.
Let $X$ be a topological space. A filter $\mathscr{F}$ on $X$ is
  said to be convergent to a point $x$ if $\mathscr{F}$ is
  finer than the neighbourhood filter of $x$. If $\mathscr{F}$
  converges to $x$ then $x$ is called a limit of $\mathscr{F}$
  and one writes $\mathscr{F} \to x$ or $x \in \lim \mathscr{F}$. One
  says that $x = \lim \mathscr{F}$ if $x$ is the only limit of
  $\mathscr{F}$. A filter basis $\mathscr{B}$ is said to be
  convergent to $x$ if the filter with the basis $\mathscr{B}$
  converges to $x$.
Let $X$ be a topological space.  The following assertions are
  equivalent:
H Every two distinct points of $X$ have distinct
  neighbourhoods.
H$'$ Every point of $X$ is the intersection of its closed
  neighbourhoods.
H$''$ The diagonal of $X \times X$ is closed.
H$'''$ There exists no filter on $X$ that converges to two
  different points.
A topological space $X$ is said to be Hausdorff or separated
  if it satisfies one (and hence all) of these conditions.
Let $f \colon X \to X'$ be a mapping [function] and $\mathscr{F}$ a
  filter on $X$. The filter on $X'$ that has images of the sets from
  $\mathscr{F}$ as a basis is called the image of $\mathscr{F}$
  relative to $f$ and will be denoted by $f(\mathscr{F})$.
If $\mathscr{B}'$ is a filter basis on $X'$, then the inverse image
  of sets from $\mathscr{B}'$ relative to $f$ satisfy condition
  FB 1. Condition FB 2 is satisfied if and only if
  no set from $\mathscr{B}'$ has an empty inverse image.
If $\mathscr{F}'$ is a filter on $X'$ and the inverse
  images of sets from $\mathscr{F}'$ are non-empty, then these images
  form, in general, only a filter basis.
Let $f \colon X \to X'$ be a mapping and $\mathscr{F}'$ a filter on
  $X'$. If the inverse images of sets from $\mathscr{F}'$ form a
  filter basis on $X$, then the filter with this basis is called the
  inverse image of $\mathscr{F}'$ relative to $f$ and will be
  denoted by $f^{-1}(\mathscr{F}')$.
Let $\mathscr{B}'$ be a filter basis on $X'$. The filter
  $\mathscr{F}'$ with basis $\mathscr{B}'$ has an inverse image
  relative to the mapping $f \colon X \to X'$ if and only if no set
  from $\mathscr{B}'$ has an empty inverse image.
The following important special case is covered by these
  considerations. Let $A$ be a subset of $X$. Relative to the
  inclusion of $A$ in $X$ every filter $\mathscr{G}$ on $A$ has an
  image on $X$ which in this case is called the extension of
  $\mathscr{G}$ to $X$. If $\mathscr{F}$ is a filter on $X$, then the
  inverse image of $\mathscr{F}$, relative to the inclusion of $A$ in
  $X$, need not exist. It exists if and only if $A$ meets every set
  from $\mathscr{F}$. This inverse image is also called the
  trace of $\mathscr{F}$ on $A$.
We introduce some more terminology and assume for this purpose that
  $Y$ is Hausdorff. The continuity of the mapping $f \colon X \to Y$
  at the point $x_0$ is equivalent to
  $$
\lim_{x\to x_0} f(x) = f(x_0),
$$
  which asserts that the image of the neighbourhood filter of $x_0$
  converges to $f(x_0)$. If the neighbourhood filter of $x_0$ induces
  a filter $\mathscr{F}$ on a subset $A$ of $X$ [it is clear that by
  this form of words he means that $\mathscr{F}$ has a trace on $A$],
  then we write
  $$
\lim_{\substack{x\to x_0\\x\in A}} f(x) = y,
$$
  when the image of $\mathscr{F}$ relative to the mapping $f|_A$
  converges to $y$. If in particular $A = X \setminus \{x_0\}$, then
  we will also write
  $$
\lim_{\substack{x\to x_0\\x\ne x_0}} f(x) = y.
$$
  [It is clear that this notation is defined if and only if $x_0$ is
  not an isolated point of $X$; and then $\mathscr{F}$ is the
  collection of all punctured neighbourhoods of $x_0$ in $X$.]

