Tricky epsilon-delta proof $$\lim_{x\to -1}\frac{x-1}{x^2-x+1}=-\frac{2}{3}$$
What I've got so far is that:
$$\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<|x-(-1)|< \delta\implies\left|\frac{x-1}{x^2-x+1}-(-\frac{2}{3})\right|<\epsilon\\
\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<| x+1|< \delta\implies\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|<\epsilon$$
How do I go about finding a value for delta from here? Thanks.

Take the case where $x\in [-2,0]$ (i.e. $\delta < 1$). Then we have the following inequality
$$\left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1|$$
by maximizing the numerator and minimizing the denominator. So set $\delta = \min(\frac{3}{5}\epsilon,1)$ and the proof step for the limit follows in both cases.
If $\epsilon > \frac{5}{3}$:
$$|x+1|<1\implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \frac{5}{3} < \epsilon$$
If $\epsilon \leq \frac{5}{3}$:
$$|x+1|<\frac{3}{5}\epsilon \implies \left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1| < \epsilon$$
