Limits, Continuity & Differentiability
Product of sec — Taylor Approximation
nta_pyq_2026_jan
Grade 12
Question:
The value of $\displaystyle\lim_{x\to0}\frac{\log_e(\sec(ex)\cdot\sec(e^2x)\cdots\sec(e^{10}x))}{e^2-e^{2\cos x}}$ is equal to
\dfrac{e^{20}-1}{2(e^2-1)}
\dfrac{e^{10}-1}{2(e^2-1)}
\dfrac{e^{10}-1}{2e^2(e^2-1)}
\dfrac{e^{20}-1}{2e^2(e^2-1)}
Step-by-Step Solution
Key Concept: Numerator: $\sum_{k=1}^{10}\log_e\sec(e^k x)$. For small $u$, $\log\sec u\approx\tfrac{u^2}{2}$. So numerator $\approx\tfrac{x^2}{2}\sum_{k=1}^{10}e^{2k}=\tfrac{x^2}{2}\cdot\tfrac{e^2(e^{20}-1)}{e^2-1}$.
Limit $=\dfrac{e^{20}-1}{2(e^2-1)}$.
Correct Answer: 1