Question

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

• A. $-1$
• B. $-\left(\frac{1}{2}\right)$
• C. $0$
• D. $1$

Answer 1

Answer: (B)

$-\left(\frac{1}{2}\right)$

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

We know that the given expression is in an indeterminate form at $x=1.$

Applying $L^{\prime }$ Hospital' s rule, we get $\underset{x\to 1}{lim}\frac{1+\log x-x}{1-2x+x^2}=\underset{x\to 1}{lim}\frac{\left(\frac{1}{x}\right)-1}{-2+2x}=\underset{x\to 1}{lim}\frac{1-x}{2x(x-1)}=\underset{x\to 1}{lim}\frac{-1}{2x}=-\frac{1}{2}.$