Can limits be defined in a more algebraic way, instead of using the completely analytic $\delta$-$\epsilon$ definition? 
Let $(X,d_X), (Y,d_Y)$ be metric spaces. Let $f:E\subseteq X\to Y$ and $a\in X$.
We say that $\lim_{x\to a}f(x)=L$ if and only if for every $\epsilon >0$ there exists a $\delta>0$ such that $d_Y(f(x),L)<\epsilon$ whenever $0<d_X(x,a)<\delta$.

This is the textbook definition you'll see in many analysis books. The notion of continuity in a metric space is most of the times defined using limits.
However, a more general (topological) definition of continuity can be stated:

$f:X\to Y$ is continuous if and only if the preimage of every open set $V\subseteq Y$ is open.


I'm looking for a nice definition like that one for limits. One that doesn't have the typical $\epsilon$-$\delta$ style.
Is such a thing possible?
 A: In some settings you can define limits in a covariant fashion, thereby making the definition considerably more intuitive: 

$\lim\limits_{x\to a}f(x)=L$ if and only if $f$ takes every point close to $a$ (in some well-defined sense) to a point close to $L$ (in the same sense).

For example, take a look at the Wikipedia article on near sets and, especially, at the paper P. Cameron, J.G. Hocking, and S.A. Naimpally, Nearness — A Better Approach to Continuity and Limits: once you have the appropriate notion of nearness, you can say that the limit of $f$ at $a$ is $L$ if $f(x)$ is near $L$ whenever $x$ is near $a$. 
Much the same sort of thing can be done in the setting of non-standard analysis: the limit of $f$ at $a$ is $L$ if the hyperreal extension $f^*$ of $f$ takes points infinitely close to $a$ to points infinitely close to $L$.
A: We can get a definition like this for limits of sequences: $a_n$ converges to $a$ if every open set containing $a$ also contains all but finitely many of the terms of $a_n$. For limits of  functions, we can say that $\lim_{x\to a} f(x)=L$ if $\lim_{n\to\infty} f(a_n)=L$ whenever $a_n$ is a sequence converging to $L$.
A: Well, once limits in $\Bbb R$ are known, we can reduce the things to sequences and limits only in $\Bbb R$:


*

*$a_n\to a$ in a metric space iff $d(a_n,a)\to 0$.

*$\lim_{x\to a} f(x)=b$ iff for all $x_n\to a$ we have $f(x_n)\to b$.

A: Suppose $f : X \to Y$ is a map between two topological spaces (not necessarily continuous); $a \in X$; and $L \in Y$.  We have that $\lim_{x\to a} f(x) = L$ if and only if whenever $U \subseteq Y$ is a neighborhood of $L$, then $f^{-1}(U)$ is a possibly punctured neighborhood of $a$.  Here, we define $V \subseteq X$ to be a possibly punctured neighborhood of $a$ if there is some open neighborhood $W$ of $a$ such that $W \setminus \{ a \} \subseteq V$ - or equivalently, if $V \cup \{ a \}$ is a neighborhood of $a$.
(Note that, just as the set of neighborhoods of $L$ forms a filter on $Y$, so also the set of possibly punctured neighborhoods of $a$ forms a filter on $X$ assuming that $a$ is not an isolated point in $X$.  This leads naturally to ways of expressing other familiar limit statements from real analysis in terms of relations between a filter on $Y$ and a filter on $X$.  For example, for $f : \mathbb{R} \to \mathbb{R}$, we can say $\lim_{x\to \infty} f(x) = \infty$ if and only if $f^{-1}(S) \in \mathscr{F}_{\infty}$  for every $S \in \mathscr{F}_\infty$, where $\mathscr{F}_\infty$ is the filter on $\mathbb{R}$ consisting of sets $S$ such that $S \supseteq [N, \infty)$ for some $N \in \mathbb{R}$.)
