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. \( e \)
• C. \( e^{-1} \)
• 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

Let's begin by analyzing the given limit: \[ \lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - \frac{x}{1 + x} \right) \right)^x. \] As \( x \to \infty \), we can simplify the expression inside the limit: \[ \frac{x}{1 + x} \to 1 \quad \text{so} \quad \frac{1}{e} - \frac{x}{1 + x} \to \frac{1}{e} - 1. \] This yields: \[ \lim_{x \to \infty} \left( \frac{e}{1 - e} \left( \frac{1}{e} - 1 \right) \right)^x. \] Now, solving for \( \alpha \), we get \( \alpha = e^{-1} \). The expression for \( \frac{\log_e \alpha}{1 + \log_e \alpha} \) becomes: \[ \frac{\log_e e^{-1}}{1 + \log_e e^{-1}} = \frac{-1}{1 - 1} = e^{-1}. \] Thus, the correct answer is \( e^{-1} \).