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$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=$
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Verified Answer
The correct answer is:
$-\frac{1}{2}$
Applying L-Hospital's rule,
$\begin{aligned}
& \lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2 x}=\lim _{x \rightarrow 1} \frac{1-x}{2 x(x-1)} \\
& \text { Again applying L-Hospital's rule, } \lim _{x \rightarrow 1} \frac{-1}{4 x-2}=-\frac{1}{2} .
\end{aligned}$
$\begin{aligned}
& \lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2 x}=\lim _{x \rightarrow 1} \frac{1-x}{2 x(x-1)} \\
& \text { Again applying L-Hospital's rule, } \lim _{x \rightarrow 1} \frac{-1}{4 x-2}=-\frac{1}{2} .
\end{aligned}$
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