Question Number 176672 by youssefelaour last updated on 24/Sep/22
Answered by Mathspace last updated on 25/Sep/22
$${we}\:{do}\:{the}\:{changement}\:{x}−\frac{\pi}{\mathrm{2}}={t} \\ $$$${so}\:{l}\left({x}\right)=\frac{\left(\mathrm{1}−{cost}\right)\left(\mathrm{1}−{cos}^{\mathrm{2}} {t}\right)....\left(\mathrm{1}−{cos}^{{n}} {t}\right)}{{sin}^{\mathrm{2}{n}} {t}} \\ $$$${but}\:\mathrm{1}−{cost}\sim\frac{{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{1}−{cos}^{\mathrm{2}} {t}=\left(\mathrm{1}−{cost}\right)\left(\mathrm{1}+{cost}\right) \\ $$$$\sim\mathrm{2}\frac{{t}^{\mathrm{2}} }{\mathrm{2}}={t}^{\mathrm{2}} \\ $$$${cost}\sim\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{cos}^{{n}} {t}\sim\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right)^{{n}} \\ $$$$\mathrm{1}−\frac{{nt}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{1}−{cos}^{{n}} {t}\sim\frac{{nt}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}}.....\frac{{nt}^{\mathrm{2}} }{\mathrm{2}}}{{t}^{\mathrm{2}{n}} } \\ $$$$=\frac{\frac{{n}!{t}^{\mathrm{2}{n}} }{\mathrm{2}^{{n}} }}{{t}^{\mathrm{2}{n}} }=\frac{{n}!}{\mathrm{2}^{{n}} }\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {f}\left({x}\right)=\frac{{n}!}{\mathrm{2}^{{n}} } \\ $$
Commented by youssefelaour last updated on 25/Sep/22
$${thanks} \\ $$