Question

Evaluate $\underset{x⟶1}{lim}(\frac{2}{1-x^2}+\frac{1}{x-1})$

Answer 1

We have,

${lim}_{x\to 1}(\frac{2}{1-x^2}+\frac{1}{x-1})$

Then,

${lim}_{x\to 1}(\frac{2}{1-x^2}+\frac{1}{x-1})$

$={lim}_{x\to 1}\left(\frac{2(x-1)+1(1-x^2)}{(1-x^2)(x-1)}\right)$

$={lim}_{x\to 1}\left(\frac{2x-2+1-x^2}{(1-x^2)(x-1)}\right)$

$={lim}_{x\to 1}\left(\frac{-x^2+2x-1}{(1-x^2)(x-1)}\right)$

$={lim}_{x\to 1}\left(\frac{-(x^2-2x+1)}{(1-x^2)(x-1)}\right)$

$={lim}_{x\to 1}\left(\frac{-{(x-1)}^2}{(1-x^2)(x-1)}\right)$

$={lim}_{x\to 1}\left(\frac{-{(x-1)}^2}{(1-x)(1+x)(x-1)}\right)$

$={lim}_{x\to 1}\left(\frac{-{(x-1)}^2}{-(x-1)(1+x)(x-1)}\right)$

$={lim}_{x\to 1}\left(\frac{1}{(1+x)}\right)$

$=\frac{1}{1+1}$

$=\frac{1}{2}$

Hence, this is the answer.