Question

The value of \[ \lim_{x\to 0}\frac{\log_e\!\big(\sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x)\big)} {e^2-e^{2\cos x}} \] is equal to:

• A. \( \dfrac{e^{10}-1}{2e^2(e^2-1)} \)
• B. \( \dfrac{e^{20}-1}{2e^2(e^2-1)} \)
• C. \( \dfrac{e^{10}-1}{2(e^2-1)} \)
• D. \( \dfrac{e^{20}-1}{2(e^2-1)} \)

Hint:

In limits involving products of trigonometric functions, convert products to sums using logarithms and then apply standard series expansions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a limit of the form \(0/0\). Using logarithmic properties and expansion formulas makes the calculation easier.
Step 2: Key Formula or Approach:
\(\lim_{x \to 0} \ln(\sec ax) = \lim_{x \to 0} -\ln(\cos ax) \approx -\ln(1 - \frac{a^2x^2}{2}) \approx \frac{a^2x^2}{2}\).
Step 3: Detailed Explanation:
The numerator is \(\sum_{k=1}^{10} \ln(\sec(e^k x)) \approx \sum_{k=1}^{10} \frac{e^{2k}x^2}{2} = \frac{x^2}{2} (e^2 + e^4 + \dots + e^{20})\).
Using GP sum: \(S = e^2 \frac{(e^2)^{10} - 1}{e^2 - 1} = \frac{e^2(e^{20} - 1)}{e^2 - 1}\).
Denominator: \(e^2 - e^{2\cos x} = e^2(1 - e^{2(\cos x - 1)})\).
Since \(\cos x - 1 \approx -x^2/2\), \(e^{2(\cos x - 1)} \approx e^{-x^2} \approx 1 - x^2\).
Denominator \(\approx e^2(1 - (1 - x^2)) = e^2 x^2\).
Limit \(L = \lim_{x \to 0} \frac{\frac{x^2}{2} \cdot \frac{e^2(e^{20} - 1)}{e^2 - 1}}{e^2 x^2} = \frac{e^{20} - 1}{2(e^2 - 1)}\).
Step 4: Final Answer:
The value is \(\frac{(e^{20} - 1)}{2(e^2 - 1)}\).