$\lim_{x\to 1^+}{\frac{1}{x^2-1}}$(with epsilon-delta) I want to prove that $\lim_{x\to 1^+}{\frac{1}{x^2-1}}=+\infty$

So I tried to define the $\delta$:

$\frac{1}{x^2-1}=\frac{1}{(x+1)(x-1)}>\frac{1}{\delta (x+1)}(\because 0<x-1<\delta)$

My problem is How to evaluate $\frac{1}{x+1}$ and How to proceed with the discussion after this.

Thanks,for any help.
 A: By definition of $\lim_{x\to 1^{+}}\dfrac{1}{x^{2}-1}=+\infty$ we have proof
$$
(\forall \epsilon>0)(\exists \delta>0)(\forall x\in \mathbb{R})
\left(( 0<\color{red}{x-1}<\delta)\implies \left(\dfrac{1}{x^{2}-1}>\color{red}{M}\right)  \right)
$$
The secret here is to go working on the expression $\left(\dfrac{1}{x^{2}-1}>\color{red}{M}\right)$ until You get the expression $\left(0<x-1<\delta\right)$.
\begin{align}
\dfrac{1}{x^{2}-1}>\color{red}{M} 
\implies &
{x^{2}-1}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{(x+1)\color{red}{(x-1)}}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{(\color{red}{(x-1)}+2)\color{red}{(x-1)}}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{\color{red}{(x-1)}^{2}+2\cdot\color{red}{(x-1)}}<\dfrac{1}{\color{red}{M} }
\\
\implies &
{\color{red}{(x-1)}^{2}+2\cdot\color{red}{(x-1)}}+1<\dfrac{1}{\color{red}{M}}+1
\\
\implies &
(\color{red}{(x-1)} +1)^{2}<\dfrac{1}{\color{red}{M}}+1
\\
\implies &
|\color{red}{(x-1)} +1|<+\sqrt{\dfrac{1}{\color{red}{M}}+1}
\\
\implies &
0<\color{red}{(x-1)} +1<+\sqrt{\dfrac{1}{\color{red}{M}}+1} 
\\
\implies &
0<\color{red}{(x-1)} <+\sqrt{\dfrac{1}{\color{red}{M}}+1} - 1
\end{align}
Now the job is with you. Show that
$$
0<x-1<\delta = \sqrt{\dfrac{1}{\color{red}{M}}+1} - 1 \mbox{ implies } \dfrac{1}{x^{2}-1}>\color{red}{M}
$$
A: If $1<x<1+\delta$ then $\frac  1{x^{2}-1} >\frac 1 {\delta^{2}+2\delta}$. Given any $M$ we want to choose $\delta$ so that $\frac 1 {\delta^{2}+2\delta}>M$ or $\delta^{2}+2\delta <\frac  1 M$. One way to get such a $\delta$ is to take $\delta <1$ so that $\delta^{2}+2\delta<3\delta$ and then we only need $3\delta <\frac 1 M$. Hence $0 <\delta <\min \{1,\frac 1 {3M}\}$ does the job.
