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?

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.