Why does factoring eliminate a hole in the limit? $$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5}  = \lim_{x\rightarrow 5} (x+5)$$
I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the $x^2-25$ into $(x-5)(x+5)$, but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the $(x-5)$, which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure."
I really want to understand this in abstract algebra terms, please elaborate. Thank you very much. 

Let's consider a simpler example first.  Consider the function $f(x) = \frac{2x}x$.  This says you take some number $x$, multiply by 2, then divide by the original number $x$.  Obviously the answer is always 2, right?  Except that when $x$ is zero, the division is forbidden and there is no answer at all. But for every $x$ except 0, we have $\frac{2x}x = 2$. In particular, for values of $x$ close to, but not equal to 0, we have $\frac{2x}x = 2$.
The function $\frac{x^2-25}{x-5}$ is similar, just a little more complicated.  Calculating $x^2-25$ always gives you the same as $(x-5)(x+5)$.  That is, if you take $x$, square it, and subtract 25, you always get the same number as if you take $x$, add 5 and subtract 5, and then multiply the two results.  So we can replace $x^2-25$ with $(x+5)(x-5)$ because they always give the same number regardless of what you start with; they are two ways of getting to the same place.  And then we see that $$\frac{x^2-25}{x-5} = \frac{(x+5)(x-5)}{x-5} = x+5$$
except that if $x-5$ happens to be zero (that is, if $x=5$) the division by zero is forbidden and we get nothing at all.  But for any other $x$ the result of $\frac{x^2-25}{x-5}$ is always exactly equal to $x+5$. In particular, for values of $x$ close to, but not equal to 5, we have $\frac{x^2-25}{x-5} = x+5 $.
The limit $$\lim_{x\to 5} \ldots$$
asks 
 what happens to some function when $x$  close to, but not exactly equal to 5.  And while this function is undefined for $x=5$, because to calculate it  you would have to divide by zero, it is perfectly well-behaved for other values of $x$, and in particular for values of $x$ close to 5.  For values of $x$ close to 5 it is equal to $x+5$, and so for values of $x$ close to 5 it is close to 10.  And that is exactly what the limit is calculating.
