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Let $X$ and $Y$ be metric spaces and let $a\in X$ be a limit point of $X$. Suppose that $f\colon X\setminus\{a\}\to Y$ is a function defined everywhere except at $a$. For $L\in Y$, we say the limit of $f(x)$ as $x$ approaches $a$ is equal to $L$, or
$\lim_{{x\to a}}f(x)=L$ 
if, for every real number $\varepsilon>0$, there exists a real number $\delta>0$ such that, whenever $x\in X$ with $0<d_{X}(x,a)<\delta$, then $d_{Y}(f(x),L)<\varepsilon$.
The formal definition of limit as given above has a well–deserved reputation for being notoriously hard for inexperienced students to master. There is no easy fix for this problem, since the concept of a limit is inherently difficult to state precisely (and indeed wasn’t even accomplished historically until the 1800’s by Cauchy, well after the development of calculus in the 1600’s by Newton and Leibniz). However, there are number of related definitions, which, taken together, may shed some light on the nature of the concept.

The notion of a limit can be generalized to mappings between arbitrary topological spaces, under some mild restrictions. In this context we say that $\lim_{{x\to a}}f(x)=L$ if $a$ is a limit point of $X$ and, for every neighborhood $V$ of $L$ (in $Y$), there is a deleted neighborhood $U$ of $a$ (in $X$) which is mapped into $V$ by $f$. One also requires that the range $Y$ be Hausdorff (or at least $T_{1}$) in order to ensure that limits, when they exist, are unique.

Let $a_{n},n\in\mathbb{N}$ be a sequence of elements in a metric space $X$. We say that $L\in X$ is the limit of the sequence, if for every $\varepsilon>0$ there exists a natural number $N$ such that $d(a_{n},L)<\varepsilon$ for all natural numbers $n>N$.
In calculus, $X$ and $Y$ are frequently taken to be Euclidean spaces $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$, in which case the distance functions $d_{X}$ and $d_{Y}$ cited above are just Euclidean distance.
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Comments
generalizations
is it possible to generalize and include limits to infinity ? and limits that equalize infinity ?
Re: generalizations
Real Variable Analysis Model of infinite cardinal numbers
Assert that
limit as x > + infinity of x exists.
Give this limit the name N.
We may, with consistency, identify N with any particular
infinite cardinal number.
I choose to identify N with AlephNull.
For any real valued continuous function f,
such that limit as x > +infinity, is + infinity,
identify f(N) with limit as x > + infinity of f(x).
Assert that this limit, f(N) exists.
For any two realvalued continued functions f and g,
such that
limit as x > + infinity of f(x) is + infinity,
and
limit as x > + infinity of g(x) is + infinity,
Define
f(N) > g(N) if and only if
limit as x > + infinity of
[ ln(ln(f(x))) / ln(ln(g(x))) ] > 1.
Define f(N) = g(N) if and only if
f(N) is not > g(N) and g(N) is not > g(N).
The operator symbol > is interchangable with the
words "greater than".
Define f(N) < g(N) if and only if g(N) > f(N).
The operator symbol < is interchangable with the
words "less than".
By this definition let's check if the number of prime numbers
is less than N.
The number of prime numbers, by extending the prime number
theorem, is N/ln(N).
limit as x > + infinity of [ln(ln( N/ln(x) ) )]/[ ln(ln(x))]
= limit as x>+ infinity of
[ln( ln(x)  ln(ln(x) ) ] / [ ln(ln(x) ]
= by the derivative test, limit as x > + infinity of
[(1/x  1/( x ln(x) )/( ln(x)  ln(ln(x)) )]/[ 1/(x ln(x))]
= limit as x > + infinity of
( ln(x)  1) / (ln(x)  ln(ln(x)) )
= by the derivative test,
limit as x > + infinity of
(1/x) / (1/x  1/(x ln(x) )
= limit as x > + infinity of
1/(1  1/ln(x))
= 1.
Theorem.
If limit as x > + infinity of f(x)/g(x) = 1,
then f(N) = g(N).
Proof.
If limit as x> + infinity of f(x)/g(x) = 1
then also,
limit as x > + infinity of g(x)/f(x) = 1.
Thus f(N) is not > g(N),
and g(N) is not > f(N).
Thus f(N) = g(N)
In this model of comparing infinities,
the number of prime positive integers is
the same as the number of all positive integers.
You may email question and comments directly to me at
kermit@polaris.net
if you wish a response sooner than I would see posts on PlanetMath.