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 $|x-1|<\delta \implies \dfrac{|x-1||2x+1|}{3|x^2+x+1|}<\epsilon$.

Here it is another way to solve it for the sake of curiosity.
To begin with, notice that
\begin{align*}
\frac{|x-1||2x+1|}{3|x^{2} + x + 1|} = \frac{|x-1||(2x - 2) + 3|}{3|x^{2} + x + 1|} \leq \frac{2|x-1|^{2} + 3|x-1|}{3|x^{2}+x+1|}
\end{align*}
Since we have that
\begin{align*}
x^{2} + x + 1 = \left(x^{2} + x + \frac{1}{4}\right) - \frac{1}{4} + 1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4} \geq \frac{3}{4} > 0 \Rightarrow 0 < \frac{1}{x^{2}+x+1} \leq\frac{4}{3}
\end{align*}
we can conclude that
\begin{align*}
\frac{|x-1||2x+1|}{3|x^{2} + x + 1|} \leq \frac{8|x-1|^{2} + 12|x-1|}{9} \leq \frac{8\delta^{2} + 12\delta}{9} = \varepsilon
\end{align*}
Thus for every $\varepsilon > 0$ there corresponds a $\displaystyle\delta = \frac{-3 + 3\sqrt{1+2\varepsilon}}{4} > 0$ such that
\begin{align*}
0 < |x - 1| \leq \delta \Rightarrow \left|\frac{x^{2}-1}{x^{3}-1} - \frac{2}{3}\right| \leq \varepsilon
\end{align*}
and we are done.
Hopefully this helps!
