Question

The value of \(\lim_{x \to 0} \frac{\ln(\sec(ex) \cdot \sec(e^2x) \dots \sec(e^{10}x))}{e^2 - e^{2\cos x}}\) is :

• A. \(\frac{1}{2} \frac{(e^2 - 1)}{(e^{20} - 1)}\)
• B. \(\frac{1}{2} \frac{(e^{20} - 1)}{(e^2 - 1)}\)
• C. \(\frac{1}{2} \frac{(e - 1)}{(e^{20} - 1)}\)
• D. None of these

Hint:

Using Taylor expansions like \( \cos x \approx 1 - x^2/2 \) is often much faster for limits involving transcendental functions than L'Hopital's rule.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Use the property \( \ln(ABC) = \ln A + \ln B + \ln C \).
Also use standard limits: \( \lim_{u \to 0} \frac{\ln(\sec u)}{u^2} = \frac{1}{2} \).
Step 2: Key Formula or Approach:
The numerator is \( \sum_{j=1}^{10} \ln(\sec(e^j x)) \).
As \( x \to 0 \), \( \ln(\sec(e^j x)) \approx \frac{(e^j x)^2}{2} \).
Step 3: Detailed Explanation:
Numerator \( \approx \frac{x^2}{2} \sum_{j=1}^{10} e^{2j} = \frac{x^2}{2} \left[ \frac{e^2(e^{20}-1)}{e^2-1} \right] \).
Denominator: \( e^2 - e^{2\cos x} = e^2(1 - e^{2\cos x - 2}) \).
As \( x \to 0 \), \( 2\cos x - 2 = 2(1 - \frac{x^2}{2}) - 2 = -x^2 \).
So, \( 1 - e^{-x^2} \approx x^2 \).
Denominator \( \approx e^2 \cdot x^2 \).
Limit \( = \frac{\frac{x^2}{2} \frac{e^2(e^{20}-1)}{e^2-1}}{e^2 x^2} = \frac{e^{20}-1}{2(e^2-1)} \).
Step 4: Final Answer:
The limit is \(\frac{1}{2} \frac{(e^{20} - 1)}{(e^2 - 1)}\).