To solve the limit:

$\lim_{x \to e} \frac{1 - \frac{x}{e}}{\frac{x}{e} - 1}$

we can apply L'Hôpital's Rule, which is used when a limit initially results in an indeterminate form such as $\frac{0}{0}$. In this case, substituting $x = e$ gives:

$\frac{1 - \frac{e}{e}}{\frac{e}{e} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$

Thus, we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $x$.

Step 1: Differentiate the Numerator and Denominator

1. Numerator: The numerator is $1 - \frac{x}{e}$. Differentiating with respect to $x$ gives: $-\frac{1}{e}$

2. Denominator: The denominator is $\frac{x}{e} - 1$. Differentiating with respect to $x$ gives: $\frac{1}{e}$

Step 2: Apply L'Hôpital's Rule

Now, we substitute these derivatives back into the limit:

$\lim_{x \to e} \frac{-\frac{1}{e}}{\frac{1}{e}} = \lim_{x \to e} -1 = -1$

Final Answer

$\lim_{x \to e} \frac{1 - \frac{x}{e}}{\frac{x}{e} - 1} = -1$

Would you like further clarification on any steps?

Here are some related questions for further exploration:

1. How does L'Hôpital's Rule work for other indeterminate forms like $\frac{\infty}{\infty}$?
2. When should you avoid using L'Hôpital's Rule in evaluating limits?
3. Can this problem be solved using algebraic manipulation instead of L'Hôpital's Rule?
4. What are some common pitfalls in applying L'Hôpital's Rule?
5. How do higher-order derivatives affect the application of L'Hôpital's Rule when the indeterminate form persists?

Tip: Always confirm that you have an indeterminate form (like $\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule.