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. 

One of definitions of $\lim_{x \to A} f(x) = B$ is:
$$\forall_{\varepsilon > 0}\exists_{\delta > 0}\forall_{0 < \left|x - A\right| < \delta}\left|f(x) - B\right| < \varepsilon$$
The intuition is that we can achieve arbitrary 'precision' (put in bounds on y axis) provided we get close enough (so we get the bounds on x axis). However the definition does not say anything about the value at the point $f(A)$ which can be undefined or have arbitrary value.
One of method of proving the limit is to find the directly $\delta(\varepsilon)$. Hence we have following formula (well defined as $x\neq 5$):
$$\forall_{0 < \left|x - 5\right| < \delta}\left|\frac{x^2-25}{x-5} - 10\right| < \epsilon$$
As $x\neq 5$ (in such case $\left|x - 5\right| = 0$) we can factor the expression out 
$$\forall_{0 < \left|x - 5\right| < \delta} \left|x + 5 - 10\right| < \varepsilon $$
$$\forall_{0 < \left|x - 5\right| < \delta} \left|x - 5 \right| < \varepsilon $$
Taking $\delta(\varepsilon) = \varepsilon$ we find that:
$$\forall_{\varepsilon > 0}\exists_{\delta > 0}\forall_{0 < \left|x - 5\right| < \delta}\left|\frac{x^2-25}{x-5} - 10\right| < \varepsilon$$
The key thing is that we don't care about value at the limit.
