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Question: \(\lim_{x \rightarrow 1}\frac{1 + \log x - x}{1 - 2x + x^{2}}\)=...

lim⁡x→11+log⁡x−x1−2x+x2\lim_{x \rightarrow 1}\frac{1 + \log x - x}{1 - 2x + x^{2}}=

A

1

B

– 1

C

0

D

−12- \frac{1}{2}

Answer

−12- \frac{1}{2}

Explanation

Solution

Applying L-Hospital’s rule,

limx→11+log⁡x−x1−2x+x2=limx→11x−1−2+2x=limx→11−x2x(x−1)\underset{x \rightarrow 1}{\text{lim}}{}\frac{1 + \log x - x}{1 - 2x + x^{2}} = \underset{x \rightarrow 1}{\text{lim}}{}\frac{\frac{1}{x} - 1}{- 2 + 2x} = \underset{x \rightarrow 1}{\text{lim}}{}\frac{1 - x}{2x(x - 1)}

Again applying L-Hospital’s rule, we get limx→1−14x−2=−12\underset{x \rightarrow 1}{\text{lim}}\frac{- 1}{4x - 2} = - \frac{1}{2}