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Question-176672




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

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