Question

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

• A. $\frac{1}{3}$
• B. $\frac{-1}{2}$
• C. $\frac{1}{2}$
• D. $\frac{-1}{3}$

Answer 1

The correct option is C $\frac{1}{2}$

given $\underset{x\to 1}{lim}\frac{1}{x-1}-\frac{2}{x^2-1}$

$\frac{1}{x-1}-\frac{2}{x^2-1}=\frac{1}{x-1}-\frac{2}{(x-1)(x+1)}=\frac{x+1-2}{(x+1)(x-1)}=\frac{1}{x+1}$

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