lim _{x rightarrow 1}left(frac{1}{ln x}-frac{ | vedclass

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(C) Let $L = \lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$.This is an indeterminate form of type $\infty - \infty$.Combining the

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$\frac{1}{2}$

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(C) Let $L = \lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$.This is an indeterminate form of type $\infty - \infty$.Combining the terms, we get $L = \lim _{x \rightarrow 1} \frac{(x-1)-\ln x}{(x-1) \ln x}$.This is of the form $\frac{0}{0}$. Applying $L'H\hat{o}pital's$ rule:$L = \lim _{x \rightarrow 1} \frac{1 - \frac{1}{x}}{\ln x + (x-1) \cdot \frac{1}{x}} = \lim _{x \rightarrow 1} \frac{\frac{x-1}{x}}{\frac{x \ln x + x - 1}{x}} = \lim _{x \rightarrow 1} \frac{x-1}{x \ln x + x - 1}$.Applying $L'H\hat{o}pital's$ rule again:$L = \lim _{x \rightarrow 1} \frac{1}{\ln x + x \cdot \frac{1}{x} + 1} = \lim _{x \rightarrow 1} \frac{1}{\ln x + 1 + 1} = \frac{1}{0 + 2} = \frac{1}{2}$.

$\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)$

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