I'd like to write an epsilon-delta proof that $$\lim\_{x\to0} \frac{1}{x^2-1} = -1$$

My strategy in doing the proof was to choose $\delta < 1$ and then bound the $|x+1|$ and $x^2$ in $$ \left|\frac{1}{x^2-1} + 1\right| = \frac{x^2}{|x+1||x-1|}$$ If my math is correct, this yields $\frac{2x^2}{|x+1||x-1|} < 2\epsilon$. Below is the proof proper; I would be grateful if someone could verify if it's valid.

"Choose $\epsilon > 0$, and let $\delta = \min({1,2\epsilon})$. Then $0 < |x| < \delta$ implies that $|\frac{1}{x^2 - 1} + 1| < \epsilon$."