lim _{x rightarrow 1} frac{log x}{1-x} = | vedclass

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Question

(B) The given limit is $\lim _{x \rightarrow 1} \frac{\log x}{1-x}$.Substituting $x = 1$,we get the indeterminate form $\frac{\log 1}{1-1} = \frac{0}{

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) The given limit is $\lim _{x ightarrow 1} \frac{\log x}{1-x}$.Substituting $x = 1$,we get the indeterminate form $\frac{\log 1}{1-1} = \frac{0}{0}$.Applying $L'	ext{Hospital's Rule}$,we differentiate the numerator and the denominator with respect to $x$:$\lim _{x ightarrow 1} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(1-x)} = \lim _{x ightarrow 1} \frac{\frac{1}{x}}{-1}$.Evaluating the limit as $x ightarrow 1$:$\frac{\frac{1}{1}}{-1} = -1$.

$\lim _{x \rightarrow 1} \frac{\log x}{1-x} = $

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