Limits, Continuity and Differentiability Part 3

Concept:Use logarithmic expansion ($\ln(1+u) \approx u$ when $u\to0$) and standard limits: $\sec u -1 \approx \frac{u^2}{2}$ and $1-\cos u \approx \frac{u^2}{2}$ as $u\to0$.The denominator is simplified using $e^A -1 \approx A$.Explanation:Write numerator as sum of logs: $\ln \sec(e x) + \ln \sec(e^2 x) + \dots + \ln \sec(e^{10} x)$.Denominator: $e^2 - e^{2\cos x} = e^2(1 - e^{2(\cos x-1)})$.As $x\to0$, $\cos x -1 \to0$, so $e^{2(\cos x-1)}-1 \approx 2(\cos x-1)$.Hence denominator $\approx - e^2 \cdot 2(\cos x-1) = 2e^2 (1-\cos x)$.Each $\ln \sec(e^k x) \approx \sec(e^k x)-1 = \frac{1-\cos(e^k x)}{\cos(e^k x)}$.Since $\cos(e^k x) \to 1$, numerator becomes $\sum_{k=1}^{10} \bigl(1-\cos(e^k x)\bigr)$.Now $1-\cos(e^k x) \approx \frac{(e^k x)^2}{2}$, and $1-\cos x \approx \frac{x^2}{2}$.So the limit $L = \sum_{k=1}^{10} \frac{(e^k x)^2/2}{2e^2 \cdot (x^2/2)} = \sum_{k=1}^{10} \frac{e^{2k}}{2e^2}$.Cancel $x^2/2$, leaving $L = \frac{1}{2e^2}(e^2 + e^4 + e^6 + \dots + e^{20})$.This is a G.P. with first term $e^2$, common ratio $e^2$, number of terms $10$. Sum $= e^2 \frac{(e^{2})^{10}-1}{e^2-1} = e^2 \frac{e^{20}-1}{e^2-1}$.Thus $L = \frac{1}{2e^2} \cdot e^2 \frac{e^{20}-1}{e^2-1} = \frac{e^{20}-1}{2(e^2-1)}$.Answer:$\frac{e^{20}-1}{2(e^2-1)}$ which corresponds to option D.