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

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^{\sqrt{\mathrm{n}}}{(1-\frac{{\mathrm{x}}^2}{\mathrm{n}})}^{\mathrm{n}}\mathrm{dx}=???$$

Answered by witcher3 last updated on 27/Oct/23

$$\mathrm{x}=\sqrt{\mathrm{n}}.\mathrm{y}$$$$={\int }_0^1{(1-{\mathrm{y}}^2)}^{\mathrm{n}}.\sqrt{\mathrm{n}}\mathrm{dy}={\mathrm{u}}_{\mathrm{n}}$$$$=\frac{1}{2}\sqrt{\mathrm{n}}.{\int }_0^1{(1-\mathrm{t})}^{\mathrm{n}}{\mathrm{t}}^{-\frac{1}{2}}\mathrm{dt}$$$${\mathrm{u}}_{\mathrm{n}}=\frac{\sqrt{\mathrm{n}}}{2}\beta (\mathrm{n}+1,\frac{1}{2})$$$$=\frac{\sqrt{\mathrm{n}}}{2}.\frac{\mathrm{\Gamma }(\mathrm{n}+1).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\mathrm{n}+\frac{3}{2})}=\frac{\sqrt{\pi }}{2}.\frac{\mathrm{\Gamma }(\mathrm{n}+1).\sqrt{\mathrm{n}}}{\mathrm{\Gamma }(\mathrm{n}+\frac{3}{2})}$$$$\mathrm{\Gamma }\left(\mathrm{z}\right)\simeq \sqrt{2\pi }.{\mathrm{z}}^{\mathrm{z}-\frac{1}{2}}{\mathrm{e}}^{-\mathrm{z}}$$$$\frac{\mathrm{\Gamma }(\mathrm{n}+1)\sqrt{\mathrm{n}}}{\mathrm{\Gamma }(\mathrm{n}+\frac{3}{2})}\sim \frac{{(\mathrm{n}+1)}^{\mathrm{n}+\frac{1}{2}}{\mathrm{e}}^{-\mathrm{n}-1}}{{(\mathrm{n}+\frac{3}{2})}^{\mathrm{n}+1}{\mathrm{e}}_{}^{-\mathrm{n}-\frac{3}{2}}}.\sqrt{\mathrm{n}}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\mathrm{n}+1}{\mathrm{n}+\frac{3}{2}}\right)}^{\mathrm{n}}.{\left(\frac{\mathrm{n}}{\mathrm{n}+\frac{3}{2}}\right)}^{\frac{1}{2}}{\mathrm{e}}^{\frac{1}{2}}$$$${\left(\frac{\mathrm{n}+1}{\mathrm{n}+\frac{3}{2}}\right)}^{\mathrm{n}}={\mathrm{e}}^{\mathrm{nln}(1-\frac{1}{2(\mathrm{n}+\frac{3}{2})})}\to {\mathrm{e}}^{-\frac{1}{2}}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{\mathrm{n}+1}{\mathrm{n}+\frac{3}{2}}\right)}^{\mathrm{n}}.{\left(\frac{\mathrm{n}}{\mathrm{n}+\frac{3}{2}}\right)}^{\frac{1}{2}}{\mathrm{e}}^{\frac{1}{2}}\to {\mathrm{e}}^{-\frac{1}{2}}.1.{\mathrm{e}}^{\frac{1}{2}}=1$$$$\Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^{\sqrt{\mathrm{n}}}{(1-\frac{{\mathrm{x}}^2}{\mathrm{n}})}^{\mathrm{n}}\mathrm{dx}=\frac{1}{2}.\sqrt{\pi }=\frac{\sqrt{\pi }}{2}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}$$$$$$

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