The value of lim_{x to 0} frac{log_{e}(sec(ex | vedclass

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(C) Let $L = \lim_{x 	o 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^{2}x)) + ... + \ln(\sec(e^{10}x))}{e^{2} - e^{2\cos x}}$.Using the expansion $\ln(\sec

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$\frac{e^{20}-1}{2(e^{2}-1)}$

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(C) Let $L = \lim_{x 	o 0} \frac{\ln(\sec(ex)) + \ln(\sec(e^{2}x)) + ... + \ln(\sec(e^{10}x))}{e^{2} - e^{2\cos x}}$.Using the expansion $\ln(\sec 	heta) \approx \frac{	heta^{2}}{2}$ as $	heta 	o 0$, the numerator becomes $\sum_{k=1}^{10} \frac{(e^{k}x)^{2}}{2} = \frac{x^{2}}{2} \sum_{k=1}^{10} e^{2k}$.The denominator is $e^{2} - e^{2\cos x} = e^{2}(1 - e^{2\cos x - 2}) \approx e^{2}(-(2\cos x - 2)) = 2e^{2}(1 - \cos x) \approx 2e^{2}(\frac{x^{2}}{2}) = e^{2}x^{2}$.Thus, $L = \lim_{x 	o 0} \frac{\frac{x^{2}}{2} \sum_{k=1}^{10} e^{2k}}{e^{2}x^{2}} = \frac{1}{2e^{2}} \sum_{k=1}^{10} (e^{2})^{k}$.Using the geometric series sum formula $\sum_{k=1}^{n} r^{k} = \frac{r(r^{n}-1)}{r-1}$, where $r = e^{2}$ and $n = 10$:$L = \frac{1}{2e^{2}} \cdot \frac{e^{2}((e^{2})^{10} - 1)}{e^{2} - 1} = \frac{e^{20} - 1}{2(e^{2} - 1)}$.

The value of $\lim_{x \to 0} \frac{\log_{e}(\sec(ex) \cdot \sec(e^{2}x) \cdot ... \cdot \sec(e^{10}x))}{e^{2} - e^{2\cos x}}$ is equal to

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