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

Question

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?

Answer 1

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,

$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=\left|\frac{(x-1)(x+2)}{\sqrt{x+2}}\right|=|x-1|\sqrt{x+2}<3\sqrt{x+2} \le 3\sqrt{\frac{\epsilon^2}{9}}=\epsilon$$