I am trying to prove that $$\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$$ and this is my approach.
Let $\epsilon>0,\delta>0\,$ so that $$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|<\epsilon\quad\text{whenever}\quad-2<x<-2+\delta$$ We have that
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=|x-1|\sqrt{x+2}$$ Let $\delta<1$ $$-2<x<-2+\delta<-1\implies-3<x-1<-2\implies2<|x-1|<3$$ so $$|x-1|\sqrt{x+2}\leq3\sqrt{x+2}<\epsilon$$ if we pick $\,\delta\leq\min\{1,\epsilon^2/9\}.$
Would this be correct?