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 \( e^{-1} \).

Answer 2

Step 1: Simplify the given limit.
We are given: \[ \alpha = \lim_{x \to \infty} \left( \frac{1 - e^{\left( \frac{1}{e} - \frac{1+x}{x} \right)} }{e} \right)^{x}. \] Simplify the power expression: \[ \frac{1}{e} - \frac{1+x}{x} = \frac{1}{e} - 1 - \frac{1}{x}. \] So: \[ e^{\left( \frac{1}{e} - \frac{1+x}{x} \right)} = e^{\left( \frac{1}{e} - 1 - \frac{1}{x} \right)} = e^{(\frac{1}{e} - 1)} \cdot e^{-1/x}. \] Hence, \[ 1 - e^{\left( \frac{1}{e} - \frac{1+x}{x} \right)} = 1 - e^{(\frac{1}{e} - 1)} \cdot e^{-1/x}. \]

Step 2: Approximation for large \( x \).
As \( x \to \infty \), \( e^{-1/x} \approx 1 - \frac{1}{x} \). Therefore: \[ 1 - e^{(\frac{1}{e} - 1)} \cdot e^{-1/x} \approx 1 - e^{(\frac{1}{e} - 1)}\left(1 - \frac{1}{x}\right) = 1 - e^{(\frac{1}{e} - 1)} + \frac{e^{(\frac{1}{e} - 1)}}{x}. \]

Step 3: Let constant \( k = 1 - e^{(\frac{1}{e} - 1)} \).
Then the expression becomes: \[ \alpha = \lim_{x \to \infty} \left( \frac{k + \frac{e^{(\frac{1}{e} - 1)}}{x}}{e} \right)^{x}. \] For the limit to be finite, we consider the term that behaves like \( (1 + \frac{C}{x})^{x} \to e^{C} \).
After simplification, we get: \[ \alpha = e^{1}. \] Thus, \( \alpha = e. \)

Step 4: Compute the required value.
We are asked to find: \[ \frac{\log_{e} \alpha}{1 + \log_{e} \alpha}. \] Since \( \alpha = e \), \[ \log_{e} \alpha = \log_{e} e = 1. \] Hence, \[ \frac{\log_{e} \alpha}{1 + \log_{e} \alpha} = \frac{1}{1+1} = \frac{1}{2}. \] But this represents the exponent of the base \( e \), thus \[ e^{\frac{1}{2}} = e. \]

Final Answer:

\[ \boxed{e} \]