If f(x) = begin{cases} frac{log_{e} x}{x-1} & | vedclass

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(A) For a function $f(x)$ to be continuous at $x=a$,the limit $\lim_{x 	o a} f(x)$ must exist and be equal to $f(a)$.Given $f(x) = \frac{\log_{e} x}{

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(A) For a function $f(x)$ to be continuous at $x=a$,the limit $\lim_{x 	o a} f(x)$ must exist and be equal to $f(a)$.Given $f(x) = \frac{\log_{e} x}{x-1}$ for $x 
eq 1$ and $f(1) = k$.We need to evaluate $\lim_{x 	o 1} \frac{\log_{e} x}{x-1}$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'$H$ôpital's Rule:$\lim_{x 	o 1} \frac{\frac{d}{dx}(\log_{e} x)}{\frac{d}{dx}(x-1)} = \lim_{x 	o 1} \frac{1/x}{1} = \frac{1}{1} = 1$.Since the function is continuous at $x=1$,we must have $k = \lim_{x 	o 1} f(x) = 1$.

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

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