f(x) = begin{cases} frac{log x}{x-1}, & text{ | vedclass

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(C) Given that the function $f(x)$ is continuous at $x=1$,the condition for continuity is $f(1) = \lim_{x 	o 1} f(x)$.From the definition of the func

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$1$

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(C) Given that the function $f(x)$ is continuous at $x=1$,the condition for continuity is $f(1) = \lim_{x 	o 1} f(x)$.From the definition of the function,$f(1) = k$.Therefore,we need to evaluate the limit: $k = \lim_{x 	o 1} \frac{\log x}{x-1}$.Substituting $x=1$,we get the indeterminate form $\frac{0}{0}$.Applying $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$:$k = \lim_{x 	o 1} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x-1)} = \lim_{x 	o 1} \frac{1/x}{1}$.Evaluating the limit as $x 	o 1$:$k = \frac{1/1}{1} = 1$.Thus,the value of $k$ is $1$.

$f(x) = \begin{cases} \frac{\log x}{x-1}, & \text{if } x \neq 1 \\ k, & \text{if } x=1 \end{cases}$ is continuous at $x=1$,then the value of $k$ is

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