Verification of epsilon-delta proof

Question

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

Answer 1

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 $\delta =\min\bigg(1,\sqrt{\dfrac{\varepsilon}{1+\varepsilon}} \bigg)=\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}$

Answer 2

Your proof is no proof at all. You just claim that $0<|x|<\min\{1,2\varepsilon\}\implies\left\vert\frac1{x^2-1}+1\right\vert<\varepsilon$ and you stop there. Besides, the claim is false, as suggested in the comments.

Answer 3

I don't understand how did you delimited with that conditions over delta, here's my proof: We want to show $$\lim_{x\to0} \frac{1}{x^2-1} = -1$$ First at all, I'm gonna show this:

1. Let $\lim_{x\to c}f(x)=L\neq 0$ then there exists $k,\delta >0$ such that if $0<|x-c|<\delta$ then $|f(x)|>k$

Proof: Let $0<\epsilon<|L|$ (note that this works for every choose of positive epsilon), since $\lim_{x\to c}f(x)=L\neq 0$ then there exists $\delta(\epsilon):=\delta >0$ such that if $0<|x-c|<\delta$ then $|f(x)-L|<\epsilon$, and then: $$\epsilon>|f(x)-L|\geq|L|-|f(x)|\Rightarrow |f(x)|>|L|-\epsilon:=k>0 \qquad \qquad \blacksquare$$

Let $\epsilon>0$, since $\lim_{x\to 0}(x^2-1)=-1$ then there exists $M,\delta_{1}(\epsilon):=\delta_{1}>0$ such that if $0<|x|<\delta_{1}$ then $|x^2-1|>M$, now take $\delta=\min\left\{\delta_{1},\sqrt{M\epsilon} \right\}$, if $0<|x|<\delta$ therefore: $$ \left|\frac{1}{x^2-1} + 1\right| = \frac{x^2}{|x^2-1|}\leq \frac{1}{M}\delta^2<\epsilon$$