Question Number 78392 by jagoll last updated on 17/Jan/20
Answered by john santu last updated on 17/Jan/20
$${let}\:{h}=\:\mathrm{cos}\:{x}\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\sqrt[{\:\:\mathrm{3}}]{\mathrm{cos}\:\mathrm{3}{x}}\:\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{4}{x}}…\:\sqrt[{{n}}]{\mathrm{cos}\:{nx}} \\ $$$${ln}\left({h}\right)=\:{ln}\left(\mathrm{cos}\:{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{cos}\:\mathrm{2}{x}\right)+\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{cos}\:\mathrm{3}{x}\right)+…+\frac{\mathrm{1}}{{n}}{ln}\left(\mathrm{cos}\:{nx}\right) \\ $$$$\frac{{dh}}{{dx}}=\frac{−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}−\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}−\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{3}{x}}−…−\frac{\mathrm{sin}\:{nx}}{\mathrm{cos}\:{nx}} \\ $$$${now}\:{we}\:{use}\:{L}'{Hopital}\:{rule} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}+\mathrm{tan}\:\mathrm{2}{x}+\mathrm{tan}\:\mathrm{3}{x}+…+\mathrm{tan}\:{nx}}{\mathrm{2}{x}}= \\ $$$$\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+{n}}{\mathrm{2}}=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$
Commented by jagoll last updated on 17/Jan/20
$${thanks}\:{sir} \\ $$