It's quite evident that $x-2$ divides $x^3-8$ given that $x\neq2$. Since I'm not interested in evaluating $\frac{x^3-8}{x-2}$ at $x=2$ I have that $\frac{x^3-8}{x-2} = x^2+2x+4$. I've used the $\delta$-$\varepsilon$ definition to prove the limit of a quadratic polynomial before but that one had real roots and this one does not. I proved the limit of the other quadratic by factoring it and then continuing on with the proof. I have not studied complex analysis and I don't belive the author of my book expects that reader to have done so either. So how can I continue the proof? Clearly the limit is $12$ but I just don't know how to prove this.
2026-03-31 17:52:21.1774979541
Computing $\lim_{x\to2}{\frac{x^3-8}{x-2}} $ via a $\delta$-$\varepsilon$ proof
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You need to prove the limit of $x^2 + 2x+4$ is $12$, and $\Big(x^2+2x+4\Big) - 12$ has real roots. $$ \Big(x^2+2x+4\Big) - 12 = x^2+2x-8 = (x-2)(x+4). $$ If we decide in advance that $\delta$ will be no bigger than $1$, then $|x-2|<\delta$ implies $1<x<3$, so $5<x+4<7$, and so $|x+4|<7$. That means $$ |(x-2)(x+4)| < 7|x-2|. $$ That should tell you what $\delta$ is, i.e. it's either $1$ or a certain thing depedning on $\varepsilon$, whichever is less.