Verification of epsilon-delta proof 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 $|\frac{1}{x^2 - 1} + 1| < \epsilon$."

Let $\varepsilon>0$ and $x\in (-1,1)$
$$\left|\frac{1}{x^2-1} + 1\right| <\varepsilon\iff \left|\frac{x^2}{x^2-1} \right|<\varepsilon \iff \left|-\frac{x^2}{1-x^2} \right|<\varepsilon \iff \frac{x^2}{1-x^2}<\varepsilon $$
$\iff x^2<\varepsilon -x^2\varepsilon \iff x^2(1+\varepsilon)<\varepsilon\iff |x|<\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}$
So $|x|<\sqrt{\dfrac{\varepsilon}{1+\varepsilon}} \implies \left|\dfrac{1}{x^2-1} + 1\right| <\varepsilon$ 
thus $\delta =\min\bigg(1,\sqrt{\dfrac{\varepsilon}{1+\varepsilon}} \bigg)=\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}$
