Question

If \[ \lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x = \alpha, \] then the value of \[ \frac{\log_e \alpha}{1 + \log_e \alpha} \] equals:

• A. \( e^{-2} \)
• B. \( e^{-1} \)
• C. \( e \)
• D. \( e^2 \)

Hint:

When solving limits with exponential functions: - Simplify the expressions first by considering the asymptotic behavior of the terms as \( x \to \infty \). - After determining the value of the limit, use it to evaluate the required expression involving logarithms or other functions.

Answer 1

The Correct Option is B

We analyze the limit: \[\lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x.\] As \( x \to \infty \), the term \( \frac{x}{1 + x} \) approaches 1. Therefore, \( \frac{1}{e} - \frac{x}{1 + x} \) approaches \( \frac{1}{e} - 1 \). The limit expression simplifies to: \[\lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - 1 \right) \right)^x.\] Solving for \( \alpha \), we find \( \alpha = e^{-1} \). The expression \( \frac{\log_e \alpha}{1 + \log_e \alpha} \) evaluates to: \[\frac{\log_e e^{-1}}{1 + \log_e e^{-1}} = \frac{-1}{1 - 1} = e^{-1}.\] Consequently, the answer is \( e^{-1} \).