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. 

Look at this example:
$$ \lim _{\text{thing} \to zero}(\dfrac{\text{Thing}}{\text{Thing}}) \cdot \text{Another thing}$$
Here our $\text{'Thing'}$ is tending to zero, but not zero, it is something real, call it a real $\text{Thing}$, which can be cancelled merrily.
