Question-72218

Question Number 72218 by aliesam last updated on 26/Oct/19

Commented by turbo msup by abdo last updated on 26/Oct/19

let A(x)=(((sinx)/x))^(1/x^2 ) ⇒ A(x)=e^((1/x^2 )ln(((sinx)/x))) we have sinx =Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)!)) =x−(x^3 /6) +o(x^5 ) ⇒((sinx)/x)=1−(x^2 /6)+o(x^4 ) ⇒ln(((sinx)/x))=ln(1−(x^2 /6)+o(x^4 )) ∼−(x^2 /6) (x→0) ⇒(1/x^2 )ln(((sinx)/x))∼−(1/6) ⇒lim_(x→0) A(x)=e^(−(1/6)) =(1/((^6 (√e))))

$${let}\:{A}\left({x}\right)=\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${A}\left({x}\right)={e}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{{sinx}}{{x}}\right)} \\ $$$${we}\:{have}\:{sinx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:+{o}\left({x}^{\mathrm{5}} \right)\:\Rightarrow\frac{{sinx}}{{x}}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{4}} \right) \\ $$$$\Rightarrow{ln}\left(\frac{{sinx}}{{x}}\right)={ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{4}} \right)\right) \\ $$$$\sim−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{{sinx}}{{x}}\right)\sim−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {A}\left({x}\right)={e}^{−\frac{\mathrm{1}}{\mathrm{6}}} =\frac{\mathrm{1}}{\left(^{\mathrm{6}} \sqrt{{e}}\right)} \\ $$

Commented by aliesam last updated on 26/Oct/19

$${god}\:{bless}\:{you} \\ $$

Commented by mathmax by abdo last updated on 26/Oct/19

$${you}\:{are}\:{welcome}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply