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Question Number 198572 by cortano12 last updated on 22/Oct/23

lim_(x→(π/(2n))) ((√((sin 2nx)/(1+cos nx)))/(4n^2 x^2 −π^2 )) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2n}}} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{sin}\:\mathrm{2nx}}{\mathrm{1}+\mathrm{cos}\:\mathrm{nx}}}}{\mathrm{4n}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} −\pi^{\mathrm{2}} }\:=? \\ $$

Answered by MM42 last updated on 22/Oct/23

(π/(2n))−x=u lim_(u→0) ((√((sin2n((π/(2n))−u))/(1+cosn((π/(2n))−u))))/((2n((π/(2n))−u)−π)(2n((π/(2n))−u)+π))) =lim_(u→0) ((√((sin(2nu))/(1+sin(nu))))/((−2nu)(2π−2nu))) =lim_(u→0) ((√((sin(2nu))/(1+sin(nu))))/((−2nu)(2π−2nu)))= { ((+∞ if u→0^− )),((−∞ if u→0^+ )) :}

$$\frac{\pi}{\mathrm{2}{n}}−{x}={u} \\ $$$${lim}_{{u}\rightarrow\mathrm{0}} \:\frac{\sqrt{\frac{{sin}\mathrm{2}{n}\left(\frac{\pi}{\mathrm{2}{n}}−{u}\right)}{\mathrm{1}+{cosn}\left(\frac{\pi}{\mathrm{2}{n}}−{u}\right)}}}{\left(\mathrm{2}{n}\left(\frac{\pi}{\mathrm{2}{n}}−{u}\right)−\pi\right)\left(\mathrm{2}{n}\left(\frac{\pi}{\mathrm{2}{n}}−{u}\right)+\pi\right)} \\ $$$$={lim}_{{u}\rightarrow\mathrm{0}} \:\frac{\sqrt{\frac{{sin}\left(\mathrm{2}{nu}\right)}{\mathrm{1}+{sin}\left({nu}\right)}}}{\left(−\mathrm{2}{nu}\right)\left(\mathrm{2}\pi−\mathrm{2}{nu}\right)} \\ $$$$={lim}_{{u}\rightarrow\mathrm{0}} \:\frac{\sqrt{\frac{{sin}\left(\mathrm{2}{nu}\right)}{\mathrm{1}+{sin}\left({nu}\right)}}}{\left(−\mathrm{2}{nu}\right)\left(\mathrm{2}\pi−\mathrm{2}{nu}\right)}=\begin{cases}{+\infty\:\:\:\:{if}\:\:\:{u}\rightarrow\mathrm{0}^{−} }\\{−\infty\:\:\:\:{if}\:\:\:\:{u}\rightarrow\mathrm{0}^{+} }\end{cases} \\ $$$$ \\ $$

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