Proof by $\epsilon - \delta$ definition that $\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$ $$\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$$
I want to know if my proof is correct. If not, please show me how to fix it, if possible.
Let $\epsilon > 0$.
$\left| \dfrac{x^2-1}{x^3-1}-\dfrac23 \right|=\left| \dfrac{3x^2-3-2x^3+2}{3x^3-3} \right|=\left| \dfrac{-2x^3+3x^2-1}{3x^3-3}\right|=\left|\dfrac{(x-1)^2(2x+1)}{3(x-1)(x^2+x+1)} \right|=\dfrac{|x-1||2x+1|}{3|x^2+x+1|}$
Assume $\delta < 1$. Then $|x-1|<1$ implies:
(i) $-2 <  0 < x < 2 \implies |x| < 2$
(ii) $-5<-3<2x+1<5 \implies |2x+1|<5$
(iii) $-3<-1<x+1<3 \implies |x+1|<3$
(iv) $|x^2+x+1| = |x(x+1)+1| \le |x||x+1|+1 < 2.3+1 = 7 $
So, make $\delta = \min \{1, \dfrac{21\epsilon}{5} \}$. Then $|x-1|<\delta \implies \dfrac{|x-1||2x+1|}{3|x^2+x+1|}<\epsilon$.

Looks good to me.
I would say, though, that if you were writing this on a homework, the first part of this answer which contains the majority of the work is simply the scratch work, and you should try to flesh out the proof on the bottom a little bit more, in the standard way for these types of proofs.
