To solve this limit using L'Hôpital's Rule, we need to verify that the expression forms an indeterminate type like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Let's proceed step-by-step:

Step 1: Analyze the Indeterminate Form

The expression is: $\lim_{x \to e} \frac{1 - \frac{x}{e}}{\frac{x}{e} - 1}$

As $x \to e$:

• The numerator approaches $1 - \frac{e}{e} = 1 - 1 = 0$.
• The denominator approaches $\frac{e}{e} - 1 = 1 - 1 = 0$.

This forms a $\frac{0}{0}$ indeterminate form, so L'Hôpital's Rule applies.

Step 2: Apply L'Hôpital's Rule (First Application)

Using L'Hôpital's Rule, we take the derivatives of the numerator and the denominator with respect to $x$:

• The derivative of the numerator $1 - \frac{x}{e}$ is $-\frac{1}{e}$.
• The derivative of the denominator $\frac{x}{e} - 1$ is $\frac{1}{e}$.

So, we get: $\lim_{x \to e} \frac{-\frac{1}{e}}{\frac{1}{e}}$

Step 3: Simplify the Result

Simplifying this expression gives: $\frac{-\frac{1}{e}}{\frac{1}{e}} = -1$

Final Answer

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

Let me know if you would like further details on each step or have additional questions.

Here are five related questions for further practice:

1. How is L'Hôpital's Rule derived, and why does it work for indeterminate forms?
2. Can you apply L'Hôpital's Rule to $\lim_{x \to 0} \frac{\sin x}{x}$?
3. What are the limitations of using L'Hôpital's Rule?
4. How would the process change if the expression was not in a $\frac{0}{0}$ indeterminate form?
5. How do you determine when to stop applying L'Hôpital's Rule in a problem?

Tip: When applying L'Hôpital's Rule, always double-check that the new form after differentiation still has an indeterminate form if you need to apply it multiple times.