lim

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

whythis 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 abigger pictureof 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.

**Answer**

First, and *by definition*, when dealing with

\lim_{x\to x_0}f(x)

we *must* assume \,f\, is defined in some neighborhood of \,x_0\, **except** , perhaps, on \,x_0\, itself, and from here that in the process of taking the limit we have the right and the duty to assume \,x\, approaches \,x_0\, in any possible way **but it is never equal to it**.

Thus, and since in our case we *always* have \,x\ne x_0=5\, during the limit process , we can algebraically cancel for the whole process.

\frac{x^2-25}{x-5}=\frac{(x+5)\color{red}{(x-5)}}{\color{red}{x-5}}=x+5\xrightarrow[x\to 5]{}10

The above process shows that the original function behaves **exactly** as the straight line \,y=x+5\, **except** at the point \,x=5\, , where there exists “a hole”, as you mention.

**Attribution***Source : Link , Question Author : Emi Matro , Answer Author : DonAntonio*