Solving a problem with the definition of limit

Question

I would like to solve this limit with the definition using $\epsilon$ and $\delta$ but I can't.

$$\lim_{x\to-1} \frac{2}{1-x^2}$$

Answer 1

I suppose you mean by $\lim_{x\rightarrow -1}\frac{2}{1-x^2}$.

First of all, the answer is the limit does not exists.

Here is the proof :

Take any $\epsilon>0$. Then for all $N\in\mathbb{N}$, by taking $0<\delta<\sqrt{2/N}$, if $0<x+1\leq\delta$, then $$\frac{2}{1-x^2}\leq\frac{2}{1-(\delta-1)^2}=\frac{2}{1-\delta^2-2\delta-1}=\frac{2}{-\delta^2+2\delta}\leq-\frac{2}{\delta^2}<-N.$$ Therefore $\frac{2}{1-x^2}$ diverges with $x\rightarrow -1+0$. (does not converge from the right side) Hence by the definition of convergence, (a function converges at a given point if it has the left and right side convergence and the left and right convergences coincide) the function does not converges at $x=-1$.

Observe that also $\lim_{x\rightarrow -1+0}\frac{2}{1-x^2}=-\infty$ and $\lim_{x\rightarrow -1-0}\frac{2}{1-x^2}=+\infty$. I tried to write as detail as possible, but please ask anything if there is something unclear.

Answer 2

Hint

Suppose $|x+1|<\frac{1}{n}$. Then $|x-1|<3$, so $|1-x^2|=|1+x||1-x|<\frac{3}{n}$. Hence $$\left|\frac{2}{1-x^2}\right|>\frac{2}{3}n$$