Question

$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}$ is equal to

(a) 1

(b) -1

(c) 0

(d) $-\frac{1}{2}$

Answer 1

$\text { } \begin{aligned}\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2} & =\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2 x} \\& =\lim _{x \rightarrow 1} \frac{1-x}{2 x(x-1)} \\& =\lim _{x \rightarrow 1}-\frac{1}{2 x}=-\frac{1}{2}\end{aligned}$

So, The correct option of this question will be (D).