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I am attempting to provide a sufficient $\epsilon$-$\delta$ proof for the following limit: $$\lim_{x\to 2}\frac{x^3-8}{x^2-4}=3.$$ Here is what I have so far: for all $\epsilon>0$, there exists $\delta>0$ such that $$\left|\frac{x^3-8}{x^2-4}-3\right|<\epsilon\text{ if }0<|x-2|<\delta.$$ However, when I simplify the expression to $$\left|\frac{x^3-8}{x^2-4}-3\right|=\left|x-3+\frac{4}{x+2}\right|$$ I am confused on what to do next.

I also know that $|x-3|\geq|x-2|-1$, but is this useful to my proof? Thanks.

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1 Answer 1

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First notice that both $x^3-8$ and $x^2-4$ have a root at $x=2$.

Moreover, we may express $x^3-8=(x-2)(x^2+2x+4)$ and $x^2-4=(x-2)(x+2)$.

Therefore, if $x\neq 2$ and $x\neq -2$, we have $$\frac{x^3-8}{x^2-4}=\dfrac{x^2+2x+4}{x+2},$$

and so $$\frac{x^3-8}{x^2-4}-3=\dfrac{x^2+2x+4}{x+2}-3=\dfrac{x^2-x-2}{x+2}=(x-2)\dfrac{x+1}{x+2}$$

We want to bound $$|x-2|\dfrac{|x+1|}{|x+2|}$$

If $|x-2|<1$ and $|x-2|<\dfrac{3\epsilon}{4}$ then $1<x<3$, so $|x+1|<4$ and $|x+2|>3$.

Thus, $\dfrac{|x+1|}{|x+2|}<\dfrac{4}{3}$ and $|x-2|\dfrac{|x+1|}{|x+2|}<\dfrac{3\epsilon}{4}\cdot\dfrac{4}{3}=\epsilon$.

For this reason, taking $\delta<\min\{1,3\epsilon/4\}$ finishes the proof.

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  • $\begingroup$ This makes a lot of sense, but I am confused why we are able to say | x - 2 | < 1. Where does the 1 come from? Is this just a choice? $\endgroup$
    – Nicholas
    Apr 13 at 19:22
  • $\begingroup$ Yes, it is just a choice. You could choose $|x-2|<2$ and then we would have $\frac{|x+1|}{|x+2|}<\frac{5}{2}$. Instead of $3\epsilon/4$ we would need $2\epsilon/5$. $\endgroup$ Apr 13 at 19:41

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