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The given limit is:

\[
\lim_{{x \to e}} \frac{\ln x - 1}{x - e}
\]

This expression suggests applying L'Hôpital's Rule since substituting \( x = e \) results in an indeterminate form \( \frac{0}{0} \). L'Hôpital's Rule allows us to differentiate the numerator and the denominator separately.

### Steps:

1. Differentiate the numerator \( \ln x - 1 \) with respect to \( x \):
   \[
   \frac{d}{dx}(\ln x - 1) = \frac{1}{x}
   \]

2. Differentiate the denominator \( x - e \) with respect to \( x \):
   \[
   \frac{d}{dx}(x - e) = 1
   \]

3. Now apply L'Hôpital's Rule:
   \[
   \lim_{{x \to e}} \frac{\ln x - 1}{x - e} = \lim_{{x \to e}} \frac{\frac{1}{x}}{1} = \frac{1}{e}
   \]

Thus, the solution is:
\[
\boxed{\frac{1}{e}}
\]

Would you like further details or have any questions?

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Here are some related questions for deeper understanding:

1. What are the conditions for applying L'Hôpital's Rule?
2. How does L'Hôpital's Rule handle higher-order indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)?
3. What are some alternative methods for solving limits involving logarithms?
4. How can you confirm the result of a limit using a numerical approach?
5. What is the significance of the natural logarithm and Euler’s number \(e\) in calculus?

**Tip:** Always check if the limit leads to an indeterminate form before applying L'Hôpital's Rule.

Evaluate the limit lim (x -> e) (ln x - 1) / (x - e)

Learn how to find the limit lim (x -> e) (ln x - 1) / (x - e) using L'Hôpital's Rule. This detailed solution walks through the steps of differentiation and substitution, explaining key calculus concepts like indeterminate forms and the application of L'Hôpital's Rule. Suitable for undergraduate calculus students.

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