# Provide an ε-δ proof for the following limit

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.

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 $$1, so $$|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.

• 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? Apr 13 at 19:22
• 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 $2\epsilon/5$. Apr 13 at 19:41