$\underset{x\to 1}{lim} \frac{1+\log x-x}{1-2x+x^2}$ equals

Solution:

We have,

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

                  [By L' Hospital's Rule] 

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