Proof verification of $\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$ using $\epsilon$-$\delta$ 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? 

Seems fine, just write it in the forward direction after working backward.
Given $\epsilon >0$, we let $\delta = \min\{ 1, \frac{\epsilon^2}{9}\}$,
then if $-2 < x < -2 + \delta$, we have $-3<x-1< -2+\delta \implies  2<|x-1|<3$.
hence,
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=\left|\frac{(x-1)(x+2)}{\sqrt{x+2}}\right|=|x-1|\sqrt{x+2}<3\sqrt{x+2} \le 3\sqrt{\frac{\epsilon^2}{9}}=\epsilon$$
