Question

Evaluate the limit: \[ L = \lim_{x \to 0} \frac{35^x - 7^x - 5^x + 1}{(e^x - e^{-x}) \ln(1 - 3x)} \]

• A. \( \frac{\ln 35}{6} \)
• B. \( \frac{\ln 35}{6} \)
• C. \( \frac{\ln 2}{6} \)
• D. \( \frac{\ln (5) \cdot \ln 7}{-6} \)

Hint:

When evaluating limits involving exponentials and logarithms, use the small \( x \) approximations: \[ a^x - 1 \approx x \ln a, \quad e^x - e^{-x} \approx 2x, \quad \ln(1 - x) \approx -x. \]

Answer 1

The Correct Option is D

Step 1: Understanding the Limit We need to evaluate the limit: \[ L = \lim_{x \to 0} \frac{35^x - 7^x - 5^x + 1}{(e^x - e^{-x}) \ln(1 - 3x)}. \]
Step 2: Approximate the Numerator Using the first-order approximation \( a^x - 1 \approx x \ln a \) for small \( x \), we get: \[ 35^x - 1 \approx x \ln 35, \quad 7^x - 1 \approx x \ln 7, \quad 5^x - 1 \approx x \ln 5. \] Thus, the numerator simplifies as: \[ 35^x - 7^x - 5^x + 1 \approx x \ln 35 - x \ln 7 - x \ln 5 = x (\ln 35 - \ln 7 - \ln 5). \] Since \( \ln 35 = \ln (7 \times 5) = \ln 7 + \ln 5 \), we conclude: \[ \ln 35 - \ln 7 - \ln 5 = 0. \] Thus, the numerator behaves as \( x^2 \) and requires a second-order approximation.
Step 3: Approximate the Denominator For small \( x \), \[ e^x - e^{-x} \approx 2x. \] Also, using \( \ln(1 - 3x) \approx -3x \), the denominator simplifies as: \[ % Option (2x)(-3x) = -6x^2. \]
Step 4: Compute the Limit Since the numerator's second-order term is: \[ (1 - 7^x)(1 - 5^x) \approx (\ln 7 \cdot x)(\ln 5 \cdot x) = x^2 \ln 7 \ln 5, \] we substitute this into the fraction: \[ L = \lim_{x \to 0} \frac{x^2 \ln 7 \ln 5}{-6x^2}. \] Canceling \( x^2 \), we obtain: \[ L = \frac{\ln 7 \ln 5}{-6}. \]
Step 5: Matching with the Options The calculated limit is \( \frac{\ln 5 \cdot \ln 7}{-6} \), which matches option (D). Final Answer: The limit evaluates to (D) \( \frac{\ln(5) \cdot \ln(7)}{-6} \).