Question:medium
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:
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:
Show Hint
In limits involving products of trigonometric functions, convert products to sums using logarithms and then apply standard series expansions.
Updated On: Mar 5, 2026
- \( \dfrac{e^{10}-1}{2e^2(e^2-1)} \)
- \( \dfrac{e^{20}-1}{2e^2(e^2-1)} \)
- \( \dfrac{e^{10}-1}{2(e^2-1)} \)
- \( \dfrac{e^{20}-1}{2(e^2-1)} \)
Show Solution
The Correct Option is D
Solution and Explanation
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