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Question Number 106505 by Ar Brandon last updated on 05/Aug/20

$$\mathrm{Show}\mathrm{that}\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left[\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{x})!{\mathrm{n}}^{\mathrm{x}}}\right]=1$$

Answered by Dwaipayan Shikari last updated on 05/Aug/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{n(n-1)....xtimes\times (n-x)!}{(n-x)!n^x}$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{n(n-1)....xtimes}{n.n....xtimes}$$$$\underset{n\to \mathrm{\infty }}{\lim }(1-\frac{1}{n})(1-\frac{2}{n})....=1$$$$$$$$$$$$$$

Commented by Ar Brandon last updated on 05/Aug/20

��Thanks. �� I arrived there while trying to demonstrate that as n and p go to infinity and zero respectively, the binomial probability distribution becomes approximately equal to the poisson distribution.

Commented by Dwaipayan Shikari last updated on 05/Aug/20

������

Commented by Dwaipayan Shikari last updated on 05/Aug/20

Are you from France?

Commented by Dwaipayan Shikari last updated on 05/Aug/20

Là nuit Bonne nuit

Commented by mathmax by abdo last updated on 06/Aug/20

$$\mathrm{here}\mathrm{x}\mathrm{is}\mathrm{real}\mathrm{not}\mathrm{integr}...!$$

Commented by Ar Brandon last updated on 06/Aug/20

����Bonne Nuit, mon ami Shikari.��

Answered by mathmax by abdo last updated on 06/Aug/20

$$\mathrm{n}!\sim {\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}\left(\mathrm{stirling}\right)$$$$(\mathrm{n}-\mathrm{x})!\sim {(\mathrm{n}-\mathrm{x})}^{\mathrm{n}-\mathrm{x}}{\mathrm{e}}^{-(\mathrm{n}-\mathrm{x})}\sqrt{2\pi (\mathrm{n}-\mathrm{x}})\Rightarrow$$$$(\mathrm{n}-\mathrm{x})!{\mathrm{n}}^{\mathrm{x}}\sim {(\mathrm{n}-\mathrm{x})}^{\mathrm{n}}{(\mathrm{n}-\mathrm{x})}^{-\mathrm{x}}{\mathrm{n}}^{\mathrm{x}}{\mathrm{e}}^{-(\mathrm{n}-\mathrm{x})}\sqrt{2\pi (\mathrm{n}-\mathrm{x})}$$$$={(\mathrm{n}-\mathrm{x})}^{\mathrm{n}}{\left(\frac{\mathrm{n}}{\mathrm{n}-\mathrm{x}}\right)}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{n}}{\mathrm{e}}^{\mathrm{x}}\sqrt{2\pi (\mathrm{n}-\mathrm{x})}\Rightarrow$$$$\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{x})!{\mathrm{n}}^{\mathrm{x}}}\sim \frac{{\mathrm{n}}^{\mathrm{n}}{\mathrm{e}}^{-\mathrm{n}}\sqrt{2\pi \mathrm{n}}}{{(\mathrm{n}-\mathrm{x})}^{\mathrm{n}}{\left(\frac{\mathrm{n}}{\mathrm{n}-\mathrm{x}}\right)}^{\mathrm{x}}{\mathrm{e}}^{-\mathrm{n}}{\mathrm{e}}^{\mathrm{x}}\sqrt{2\pi (\mathrm{n}-\mathrm{x})}}$$$$={\left(\frac{\mathrm{n}}{\mathrm{n}-\mathrm{x}}\right)}^{\mathrm{n}}.{\left(\frac{\mathrm{n}-\mathrm{x}}{\mathrm{n}}\right)}^{-\mathrm{x}}{\mathrm{e}}^{-\mathrm{x}}\sqrt{\frac{2\pi \mathrm{n}}{2\pi (\mathrm{n}-\mathrm{x})}}\mathrm{but}$$$${\left(\frac{\mathrm{n}-\mathrm{x}}{\mathrm{n}}\right)}^{\mathrm{x}}={(1-\frac{\mathrm{x}}{\mathrm{n}})}^{\mathrm{x}}\to 1\mathrm{and}{\left(\frac{\mathrm{n}}{\mathrm{n}-\mathrm{x}}\right)}^{\mathrm{n}}={\left(\frac{\mathrm{n}-\mathrm{x}}{\mathrm{n}}\right)}^{-\mathrm{x}}$$$$={(1-\frac{\mathrm{x}}{\mathrm{n}})}^{-\mathrm{n}}={\mathrm{e}}^{-\mathrm{nln}(1-\frac{\mathrm{x}}{\mathrm{n}})}\to {\mathrm{e}}^{\mathrm{x}}\mathrm{and}\sqrt{\frac{\mathrm{n}}{\mathrm{n}-\mathrm{x}}}\to 1\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{x})!{\mathrm{n}}^{\mathrm{x}}}=1$$$$$$

Commented by Ar Brandon last updated on 06/Aug/20

Thanks Sir. Sorry I didn't precise that. In my case x is a natural number. But still arrived at thesame result. Thanks��

Commented by Dwaipayan Shikari last updated on 06/Aug/20

Thanking you for your corrections

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