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Question Number 119356 by Lordose last updated on 23/Oct/20

Commented by Dwaipayan Shikari last updated on 23/Oct/20

lim_(x→∞) (1/x)log((x^x /(x!))) =(1/x).log(((x^x e^x )/(x^x .(√(2πx)))))(Stirling′s approximation lim_(n→∞) n!=((n/e))^n (√(2πn)) =(1/x)log(e^x )−(1/x)log((√(2πx))) =1−(1/(2x))log(x)−(1/(2x))log(2π) lim_(x→∞) =1−(1/2).((logx)/x) (lim_(x→∞) ((logx)/x)=−((log((1/x)))/x)=−(1/x)+(1/(2x^2 ))+...=0) =1

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}{log}\left(\frac{{x}^{{x}} }{{x}!}\right) \\ $$$$=\frac{\mathrm{1}}{{x}}.{log}\left(\frac{{x}^{{x}} {e}^{{x}} }{{x}^{{x}} .\sqrt{\mathrm{2}\pi{x}}}\right)\left({Stirling}'{s}\:{approximation}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}!=\left(\frac{{n}}{{e}}\right)^{{n}} \sqrt{\mathrm{2}\pi{n}}\right. \\ $$$$=\frac{\mathrm{1}}{{x}}{log}\left({e}^{{x}} \right)−\frac{\mathrm{1}}{{x}}{log}\left(\sqrt{\mathrm{2}\pi{x}}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}{log}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}{x}}{log}\left(\mathrm{2}\pi\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{{logx}}{{x}}\:\:\:\:\:\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{logx}}{{x}}=−\frac{{log}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}=−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }+...=\mathrm{0}\right) \\ $$$$=\mathrm{1} \\ $$

Commented by Lordose last updated on 23/Oct/20

Very nice sir

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{sir} \\ $$

Answered by mathmax by abdo last updated on 23/Oct/20

we have x! ∼ x^x e^(−x) (√(2πx))⇒(x^x /(x!)) ∼(e^x /(√(2πx))) ⇒ ((ln((x^x /(x!))))/x)∼ ((ln((e^x /(√(2πx)))))/x) =((x−ln((√(2πx))))/x) =1−((ln(2πx))/(2x)) →1 ⇒ lim_(x→+∞) ((ln((x^x /(x!))))/x) =1 ⇒

$$\mathrm{we}\:\mathrm{have}\:\mathrm{x}!\:\sim\:\mathrm{x}^{\mathrm{x}} \mathrm{e}^{−\mathrm{x}} \sqrt{\mathrm{2}\pi\mathrm{x}}\Rightarrow\frac{\mathrm{x}^{\mathrm{x}} }{\mathrm{x}!}\:\sim\frac{\mathrm{e}^{\mathrm{x}} }{\sqrt{\mathrm{2}\pi\mathrm{x}}}\:\Rightarrow \\ $$$$\frac{\mathrm{ln}\left(\frac{\mathrm{x}^{\mathrm{x}} }{\mathrm{x}!}\right)}{\mathrm{x}}\sim\:\frac{\mathrm{ln}\left(\frac{\mathrm{e}^{\mathrm{x}} }{\sqrt{\mathrm{2}\pi\mathrm{x}}}\right)}{\mathrm{x}}\:=\frac{\mathrm{x}−\mathrm{ln}\left(\sqrt{\mathrm{2}\pi\mathrm{x}}\right)}{\mathrm{x}}\:=\mathrm{1}−\frac{\mathrm{ln}\left(\mathrm{2}\pi\mathrm{x}\right)}{\mathrm{2x}}\:\rightarrow\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \:\frac{\mathrm{ln}\left(\frac{\mathrm{x}^{\mathrm{x}} }{\mathrm{x}!}\right)}{\mathrm{x}}\:=\mathrm{1} \\ $$$$\Rightarrow \\ $$

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