localization of rings and polynomial functions Let $f$ and $g$ be two polynomials (polynomial functions in $n$ variables); if in some localization of the ring $k[X_1,\ldots, X_n]$ exists the class $\frac{f}{g}$, it defines in a unique way  the function  $p\mapsto f(p){(g(p))}^{-1}$. Infact  $\frac{f}{g}=\frac{f'}{g'}$ as equivalence classes if and only if they define the same function. The problem is the following: suppose that $\frac{f}{g}=\frac{f'}{g'}$ but there is a point $q$ such that $g(q)\neq0$ and $g'(q)=0$, then the function $p\mapsto \frac{f'(p)}{q'(p)}$ isn't well defined. For example $\frac{X^2}{X}=X$ in $k[X]_X$ (localizaztion of the ring $K[X]$) so the two associated functions should be the same, but clearly in $0$ there is some problem. In some textbooks it seems that the two functions must be considered however the same.
 A: Suppose $V\subset \mathbb A^N_k $ is an affine irreducible algebraic variety over an algebraically closed field $k$.
Associated to $V$  are the ring of regular functions $\mathcal O(V)$ and its fraction field $Rat(V)= Frac(\mathcal O(V)$.
The ring $\mathcal O(V)$ is exactly the ring of regular functions $V\to k$, the functions algebraic geometry is concerned with.
But what you have to keep in mind is that $Rat(V)$ does not consist of functions : it is a field formally constructed from $\mathcal O(V)$ by algebraic wizardry, period.
However to each $\phi \in Rat(V)$, you can associate a non-empty  open subset $\emptyset \neq U\subset V$ and a regular function $\tilde {\phi}:U\to k$.  The relation between $\phi$ and $\tilde {\phi}$ is the following:  

For every $p\in U$ there exist $f,g\in \mathcal O(V)$ with $g(p)\neq 0$ and $\tilde {\phi}(p)=\frac {f(p)}{g(p)}\in k$ 

Note carefully that $f$ and $g$ vary according to the choice of $p\in U$ see this recent question 
The set $U$ is the set of all $p\in V$ such that $\phi\in \mathcal O_{V,p}$ . 
In your example the function $\tilde {\phi}$ associated to $ \phi=\frac {X^2}{X}$ is exactly the function $X\in \mathcal O (V)$ since $\frac {X^2}{X}=X\in Rat(V)$ (and of course $U$ is the whole of $V$) .  
And please, please, please, never even think of pronouncing the words "limit" or "real value" or "L'Hospital rule": leave them to our Analysis friends  ...
