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How would you "remove the discontinuity" of f? in f(x) = (x^3-8)/(x^2-4)

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Restricting the domain to \[\mathbb{R}-\{-2,2\} \]
well I know that the bottom cannot be 2 but I don't understand how to go past there. I've looked up that you factor the top and the bottom but that doesn't help either. I unfortunately had to miss this lecture today and the book is almost as confusing as the proff.
Okay, that's a division, so your denominator can't be 0. You have to looik to ALL the cases in which your denominator is equal to 0 SO \[x^2-4=0\leftarrow \rightarrow (x+2)(x-2)=0\] Clearer?

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yeah. so the discontinuity is (-2,2) that's where the point would be on the graph?
it ask it in the book how would you define f(2) in order to make f continuous at 2 it would be f(2) = -2
no no no, that are the values of X in which your function is not defined. that means you don't have f(-2) nor f(2) To make it continuos you'll have to use continuity definition a function is continue for x=a if \[\lim_{x \rightarrow a}f(x)=f(a)\]
ok so how would you use that function you wouldn't plug it in? sorry I'm also trying to look through the book as you explain as well
ok in the book it shows an example right after the where it talks about the continuity definition and what it does it shows you how to find what the domain is for the bottom then it shows putting in what x approaches to like in this one lim x approaches -2 so they sub in the -2 and that's all it shows.

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