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Question Number 167120 by DAVONG last updated on 07/Mar/22

Answered by TheSupreme last updated on 07/Mar/22

((sin((π/2)(1−cos(x))))/(1−cos(x)))=(π/2)((1−cos(x))/(1−cos(x)))=(π/2)

$$\frac{{sin}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)}{\mathrm{1}−{cos}\left({x}\right)}=\frac{\pi}{\mathrm{2}}\frac{\mathrm{1}−{cos}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}=\frac{\pi}{\mathrm{2}} \\ $$

Answered by Mathspace last updated on 07/Mar/22

cosx∼1−(x^2 /2) ⇒(π/2)cosx∼(π/2)−(π/4)x^2   ⇒cos((π/2)cosx)∼cos((π/2)−(π/4)x^2 )  sin((π/4)x^2 )∼(π/4)x^2   and  1−cosx∼(x^2 /2) ⇒  l(x)∼(((π/4)x^2 )/((1/2)x^2 )) ⇒lim_(x→0)   l(x)=((2π)/4)=(π/2)

$${cosx}\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\frac{\pi}{\mathrm{2}}{cosx}\sim\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{cos}\left(\frac{\pi}{\mathrm{2}}{cosx}\right)\sim{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}{x}^{\mathrm{2}} \right) \\ $$$${sin}\left(\frac{\pi}{\mathrm{4}}{x}^{\mathrm{2}} \right)\sim\frac{\pi}{\mathrm{4}}{x}^{\mathrm{2}} \:\:{and} \\ $$$$\mathrm{1}−{cosx}\sim\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$${l}\left({x}\right)\sim\frac{\frac{\pi}{\mathrm{4}}{x}^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{l}\left({x}\right)=\frac{\mathrm{2}\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}} \\ $$

Answered by kapoorshah last updated on 07/Mar/22

L′Hopital       lim_(x→0)  ((−(π/2) . sin x . (−sin ((π/2) cos x)))/(sin x))  = (π/2) . sin ((π/2) cos 0)  = (π/2)

$${L}'{Hopital} \\ $$$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\frac{\pi}{\mathrm{2}}\:.\:\mathrm{sin}\:{x}\:.\:\left(−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\:\mathrm{cos}\:{x}\right)\right)}{\mathrm{sin}\:{x}} \\ $$$$=\:\frac{\pi}{\mathrm{2}}\:.\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{0}\right) \\ $$$$=\:\frac{\pi}{\mathrm{2}} \\ $$

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