Why does factoring eliminate a hole in the limit?

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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.

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

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