Question-78392

Question Number 78392 by jagoll last updated on 17/Jan/20

Answered by john santu last updated on 17/Jan/20

let h= cos x(√(cos 2x))((cos 3x))^(1/( 3)) ((cos 4x))^(1/4) ... ((cos nx))^(1/n) ln(h)= ln(cos x)+(1/2)ln(cos 2x)+(1/3)ln(cos 3x)+...+(1/n)ln(cos nx) (dh/dx)=((−sin x)/(cos x))−((sin 2x)/(cos 2x))−((sin 3x)/(cos 3x))−...−((sin nx)/(cos nx)) now we use L′Hopital rule lim_(x→0) ((tan x+tan 2x+tan 3x+...+tan nx)/(2x))= ((1+2+3+4+...+n)/2)= ((n(n+1))/4)

$${let}\:{h}=\:\mathrm{cos}\:{x}\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\sqrt[{\:\:\mathrm{3}}]{\mathrm{cos}\:\mathrm{3}{x}}\:\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{4}{x}}…\:\sqrt[{{n}}]{\mathrm{cos}\:{nx}} \\ $$$${ln}\left({h}\right)=\:{ln}\left(\mathrm{cos}\:{x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{cos}\:\mathrm{2}{x}\right)+\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{cos}\:\mathrm{3}{x}\right)+…+\frac{\mathrm{1}}{{n}}{ln}\left(\mathrm{cos}\:{nx}\right) \\ $$$$\frac{{dh}}{{dx}}=\frac{−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}−\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}−\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{3}{x}}−…−\frac{\mathrm{sin}\:{nx}}{\mathrm{cos}\:{nx}} \\ $$$${now}\:{we}\:{use}\:{L}'{Hopital}\:{rule} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}+\mathrm{tan}\:\mathrm{2}{x}+\mathrm{tan}\:\mathrm{3}{x}+…+\mathrm{tan}\:{nx}}{\mathrm{2}{x}}= \\ $$$$\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+{n}}{\mathrm{2}}=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$

Commented by jagoll last updated on 17/Jan/20

$${thanks}\:{sir} \\ $$

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