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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (2 + x) - \sin (2 - x)}}{x} = $

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} (\text{cosec}\,x - \cot x)$

$\lim _{x \rightarrow 0} x^3 \left\{ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right\} = $

The value of $\lim_{x \to 3} \frac{[x] - 3}{x - 3}$, where $[\cdot]$ denotes the greatest integer function, is...

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module