The value of [_x0_e[(e^x)(e^2x)(e^10x)]e^2-e^2 x] is equal to

Question

The value of
$$\lim_{x\to0}\frac{\log_e\left[\sec(e^x)\cdot\sec(e^{2x})\cdots\sec(e^{10x})\right]}{e^2-e^{2\cos x}}$$
is equal to

Accepted Answer

Write $L=\displaystyle\lim_{x\to0}\frac{\sum_{k=1}^{10}\ln\sec(e^{kx})}{e^2-e^{2\cos x}}$. Using $\ln(\sec\theta)\approx\sec\theta-1$ as $\theta\to0$ and $\sec\theta-1\to0$, and $e^{2(\cos x-1)}-1\approx2(\cos x-1)$ as $x\to0$:

$$L=\lim_{x\to0}\frac{\sum_{k=1}^{10}(\sec(e^{kx})-1)}{-e^2\cdot 2(\cos x-1)}=\lim_{x\to0}\sum_{k=1}^{10}\frac{1-\cos(e^{kx})}{2e^2(1-\cos x)}$$

Using $1-\cos\theta\approx\theta^2/2$ for small $\theta$ (with $e^{kx}\to0$, $x\to0$):

$$L=\sum_{k=1}^{10}\lim_{x\to0}\frac{(e^{kx})^2/2}{2e^2\cdot x^2/2}=\sum_{k=1}^{10}\frac{e^{2k}}{2e^2}=\frac{1}{2e^2}\left(e^2+e^4+\cdots+e^{20}\right)$$

This is a geometric series with first term $e^2$, ratio $e^2$, $10$ terms:
$$L=\frac{1}{2e^2}\cdot e^2\cdot\frac{(e^2)^{10}-1}{e^2-1}=\frac{e^{20}-1}{2(e^2-1)}$$

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