$f:(-1,1)\rightarrow \mathbb{R}, \lim_{x\rightarrow 1} \frac{x+1}{x^2-1}$. Does $f$ have a limit at 1?

Question

$f:(-1,1)\rightarrow \mathbb{R}, \lim_{x\rightarrow 1} \frac{x+1}{x^2-1}$. Does $f$ have a limit at 1?

So, my assumption is obviously that this DNE, because after manipulation and viewing graphically, we see that it does not.

I started out my proof, trying to go by the negation of a limit definition.

$\exists \epsilon>0, \forall \delta>0, 0<|x-x_o|<\delta \Rightarrow |f(x)-L|\geq\epsilon$.

Where do I go from here? I am having a hard time moving forward with this. I tried towards contradiction, but this will not work algebraically, and also I am a bit uncomfortable doing this sequentially. Just looking for some rationale to help me get through this.

Answer 1

or we get $$\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}$$ for $x\neq 1$ and $x\ne -1$

Answer 2

Assume it does, say, $L$, then for some $\delta>0$ we have if $x\in(1-\delta,1)$ then $|(x+1)/(x^{2}-1)-L|<1$, so for all such $x\in(1-\delta,1)$, $|1/(x-1)-L|<1$, or $1/(1-x)<1+|L|$. Since $1/(1-x)\rightarrow\infty$ as $x\rightarrow 1^{-}$, so $\infty\leq 1+|L|$ as we take limit both sides as $x\rightarrow 1^{-}$.