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Question Number 35736 by rahul 19 last updated on 22/May/18

$$\underset{x\to 0}{\lim }\frac{\ln (\sec \left(ex\right)\sec \left(e^{2}x\right)......\sec \left(e^{50}x\right))}{e^2-e^{2\cos x}}=?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/May/18

$$thislimitis\left(\frac{0}{0}\right)form$$$$=\frac{lim}{x\to 0}\frac{lnsec\left(ex\right)+lnsec\left(e^{2}x\right)+...+lnsec\left(e^{50}x\right)}{e^2-e^{2cosx}}$$$$usingl,hospitalrule$$$$=\frac{lim}{x\to 0}\frac{tan\left(ex\right)+tan\left(e^{2}x\right)+..+tan\left(e^{50}x\right)}{0-e^{2cosx}\times (-2sinx)}$$$$deviding{N}_{r}and{D}_{r}byx$$$$=\frac{lim}{x\to 0}\frac{e\frac{tan\left(ex\right)}{ex}+e^2\frac{tan\left(e^{2}x\right)}{e^{2}x}+..+e^{50}\frac{tan\left(e^{50}x\right)}{e^{50}x}}{2e^{2cosx}\times \frac{sinx}{x}}$$$$=\frac{e+e^2+e^3+...+e^{50}}{2e^2}$$$$=\frac{e\left(\frac{e^{50}-1}{e-1}\right)}{2e^2}$$$$=\frac{1}{2}\times \frac{1}{e}\times \frac{e^{50}-1}{e-1}$$

Commented by rahul 19 last updated on 25/May/18

Thank You Sir . There is a slight error in 2nd line .�� It will be e*tan(ex) ,......

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