Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 199084 by universe last updated on 27/Oct/23

lim_(n→∞) ∫_0 ^(√n) (1−(x^2 /n))^n dx = ???

$$\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{n}}} \:\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}}\right)^{\mathrm{n}} \mathrm{dx}\:\:\:=\:\:\:\:??? \\ $$

Answered by witcher3 last updated on 27/Oct/23

x=(√n).y =∫_0 ^1 (1−y^2 )^n .(√n)dy=u_n =(1/2)(√n).∫_0 ^1 (1−t)^n t^(−(1/2)) dt u_n =((√n)/2)β(n+1,(1/2)) =((√n)/2).((Γ(n+1).Γ((1/2)))/(Γ(n+(3/2))))=((√π)/2).((Γ(n+1).(√n))/(Γ(n+(3/2)))) Γ(z)≃(√(2π)).z^(z−(1/2)) e^(−z) ((Γ(n+1)(√n))/(Γ(n+(3/2))))∼(((n+1)^(n+(1/2)) e^(−n−1) )/((n+(3/2))^(n+1) e_ ^(−n−(3/2)) )).(√n) lim_(n→∞) (((n+1)/(n+(3/2))))^n .((n/(n+(3/2))))^(1/2) e^(1/2) (((n+1)/(n+(3/2))))^n =e^(nln(1−(1/(2(n+(3/2)))))) →e^(−(1/2)) lim_(n→∞) (((n+1)/(n+(3/2))))^n .((n/(n+(3/2))))^(1/2) e^(1/2) →e^(−(1/2)) .1.e^(1/2) =1 ⇒lim_(n→∞) ∫_0 ^(√n) (1−(x^2 /n))^n dx=(1/2).(√π)=((√π)/2)=∫_0 ^∞ e^(−x^2 ) dx

$$\mathrm{x}=\sqrt{\mathrm{n}}.\mathrm{y} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{n}} .\sqrt{\mathrm{n}}\mathrm{dy}=\mathrm{u}_{\mathrm{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{n}}.\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{n}} \mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dt} \\ $$$$\mathrm{u}_{\mathrm{n}} =\frac{\sqrt{\mathrm{n}}}{\mathrm{2}}\beta\left(\mathrm{n}+\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\sqrt{\mathrm{n}}}{\mathrm{2}}.\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right).\sqrt{\mathrm{n}}}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left(\mathrm{z}\right)\simeq\sqrt{\mathrm{2}\pi}.\mathrm{z}^{\mathrm{z}−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\mathrm{z}} \\ $$$$\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)\sqrt{\mathrm{n}}}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\sim\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\mathrm{n}−\mathrm{1}} }{\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{e}_{} ^{−\mathrm{n}−\frac{\mathrm{3}}{\mathrm{2}}} }.\sqrt{\mathrm{n}} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}}\right)^{\mathrm{n}} .\left(\frac{\mathrm{n}}{\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}}\right)^{\mathrm{n}} =\mathrm{e}^{\mathrm{nln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\right)} \rightarrow\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}}\right)^{\mathrm{n}} .\left(\frac{\mathrm{n}}{\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \rightarrow\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} .\mathrm{1}.\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{1} \\ $$$$\Rightarrow\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{n}}} \left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}}\right)^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}.\sqrt{\pi}=\frac{\sqrt{\pi}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com