Use the $\epsilon$-$\delta$ definition to prove
$$\lim_{x \to -2} \frac{x^2+x+1}{3x+3}=-1$$
Rough Work:
$$|f(x)-L| = \left| \frac{x^2+x+1}{3x+3}+1\right|= \left| \frac{x^2+4x+4}{3x+3}\right|=\frac{|x+2|^2}{|3x+3|}$$
If we restrict
$$|x+2| < 0.5$$
then
$$-0.5 < x+2 < 0.5$$ $$\implies -4.5 < 3x+3 < -1.5$$
so
$$|3x+3|>1.5$$
meaning
$$\frac{|x+2|^2}{|3x+3|} < \frac{|x+2|^2}{\frac 32}$$ $$\implies \frac{|x+2|^2}{|3x+3|} < \frac{2|x+2|^2}{3}$$ (*)
suggesting we take
$$\delta = \sqrt{\frac 32 \epsilon}$$
Proof:
Let $\epsilon > 0$.
Set $\delta = $ min $\left\{ 0.5, \sqrt{\frac 32 \epsilon} \right\}$.
Assume $|x+2| < \delta$.
$$|f(x)-L| = \frac{|x+2|^2}{|3x+3|}$$
Because $|x+2| < 0.5$, by (*),
$$\frac{|x+2|^2}{|3x+3|} < \frac{2|x+2|^2}{3}$$
And because $|x+2| < \sqrt{\frac 32 \epsilon}$,
$$\frac{2|x+2|^2}{3} < \frac{2\sqrt{\frac 32 \epsilon}^2}{3}$$ $$\implies \frac{2|x+2|^2}{3} < \epsilon$$