$\lim_{x\to 1^+}{\frac{1}{x^2-1}}$(with epsilon-delta)

Question

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.

Answer 1

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 $$ 0<x-1<\delta = \sqrt{\dfrac{1}{\color{red}{M}}+1} - 1 \mbox{ implies } \dfrac{1}{x^{2}-1}>\color{red}{M} $$

Answer 2

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.