$\underset{x\to 1}{\lim }\left(\frac{{\displaystyle 2}}{{\displaystyle 1-x^2}}+\frac{{\displaystyle 1}}{{\displaystyle x-1}}\right)$is equal to

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

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

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

$=\underset{x\to 1}{\lim }\left(\frac{{\displaystyle (1-x)(x-1)}}{{\displaystyle (1-x)(1+x)(x-1)}}\right)$

$=\underset{x\to 1}{\lim }\frac{1}{x+1}$

= $\frac{1}{2}$