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

Concept: For small \( t \), \[ \sec t = 1+\frac{t^2}{2}+o(t^2), \quad \log(\sec t)=\frac{t^2}{2}+o(t^2) \] Also, standard limits and series expansions are used.
Step 1: Simplify the numerator \[ \log_e\!\big(\sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x)\big) = \sum_{k=1}^{10}\log(\sec(e^k x)) \] For small \( x \), \[ \log(\sec(e^k x)) \sim \frac{(e^k x)^2}{2} \] Hence, \[ \text{Numerator} \sim \frac{x^2}{2}\sum_{k=1}^{10} e^{2k} \]
Step 2: Evaluate the geometric sum \[ \sum_{k=1}^{10} e^{2k} = e^2\frac{e^{20}-1}{e^2-1} \] Thus, \[ \text{Numerator} \sim \frac{x^2}{2}\, e^2\frac{e^{20}-1}{e^2-1} \]
Step 3: Simplify the denominator For small \( x \), \[ \cos x = 1-\frac{x^2}{2}+o(x^2) \Rightarrow e^{2\cos x} = e^{2-x^2} = e^2(1-x^2+o(x^2)) \] So, \[ e^2-e^{2\cos x} \sim e^2x^2 \]
Step 4: Take the limit \[ \lim_{x\to 0} \frac{\frac{x^2}{2}\, e^2\frac{e^{20}-1}{e^2-1}} {e^2x^2} = \frac{e^{20}-1}{2(e^2-1)} \]