c) \( \lim _{x \rightarrow e} \frac{\ln x-1}{x-e} \)

To find the limit \( \lim_{x \rightarrow e} \frac{\ln x - 1}{x - e} \), we can begin by substituting \( x = e \) directly: \[ \frac{\ln e - 1}{e - e} = \frac{1 - 1}{0} = \frac{0}{0} \] Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which tells us to differentiate the numerator and the denominator. Differentiating the numerator: \[ \frac{d}{dx}(\ln x - 1) = \frac{1}{x} \] Differentiating the denominator: \[ \frac{d}{dx}(x - e) = 1 \] Now, we can rewrite the limit: \[ \lim_{x \rightarrow e} \frac{\ln x - 1}{x - e} = \lim_{x \rightarrow e} \frac{\frac{1}{x}}{1} = \lim_{x \rightarrow e} \frac{1}{x} = \frac{1}{e} \] Thus, the final answer is: \[ \lim_{x \rightarrow e} \frac{\ln x - 1}{x - e} = \frac{1}{e} \]