Q:

Prove the limit $\lim_{x\rightarrow-1^{+}} = \frac{1}{x^{^{2}} -1}$ exists.

For each of the following, use definitions (rather than limit theorems) to prove that the limit exists. Identify the limit in each case. 
(c) $\lim_{x\rightarrow-1^{+}} = \frac{1}{x^{^{2}} -1}$
Proof:  By definition the function f(x) is said to converge to infinity as x → a if and only if there is an open interval I containing and given a real M, there is an δ > 0 such that 0 < |x - a| < δ implies f(x) > M, in which case f(x) approaches infinity as x → a.
Let L = infinity, and suppose ε > 0. And suppose M > 0. Then there is an δ > 0 such that |x - (- 1) | < ε . Then choose M = 
Can someone please help me prove the limit exists. I don't know how to continue.
Please, I would really appreciate it. Thank you.

A:

For any $\;M\in\Bbb R^+\;$ and $\;x>-1\;$ (but very close to $\;-1\;$)
$$\frac1{|x^2-1|}>M\iff x+1<\frac1{M|x-1|}<\frac1{2M}\implies$$
since we can make sure that $\;|x-1|>\frac32\iff\frac1{|x-1|}<\frac23\;$ , so  we can choose $\;\delta_M:=\frac2{3M}\;$ , and thus:
$$x+1<\delta_M\implies\left|\frac1{x^2-1}\right|>M$$
and the above proves
$$\lim_{x\to -1^+}\frac1{x^2-1}=-\infty$$
since $\;x<-1\implies x^2-1<0\;$

