Question

The value of $\underset{x\to 1}{lim}\frac{\ln x-x+1}{1-2x+x^2}$ equals to

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

Answer 1

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

Clearly form of the limit is, $\frac{0}{0}$
Thus using L -Hospital's rule,

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

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

$=-\frac{1}{2}$