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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

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x}log\left(\frac{x^x}{x!}\right)$$$$=\frac{1}{x}.log\left(\frac{x^x{e}^x}{x^x.\sqrt{2\pi x}}\right)({Stirling}^{\prime }sapproximation\underset{n\to \mathrm{\infty }}{\lim }n!={\left(\frac{n}{e}\right)}^n\sqrt{2\pi n}$$$$=\frac{1}{x}log\left(e^x\right)-\frac{1}{x}log\left(\sqrt{2\pi x}\right)$$$$=1-\frac{1}{2x}log\left(x\right)-\frac{1}{2x}log\left(2\pi \right)$$$$\underset{x\to \mathrm{\infty }}{\lim }=1-\frac{1}{2}.\frac{logx}{x}(\underset{x\to \mathrm{\infty }}{\lim }\frac{logx}{x}=-\frac{log\left(\frac{1}{x}\right)}{x}=-\frac{1}{x}+\frac{1}{2x^2}+...=0)$$$$=1$$

Commented by Lordose last updated on 23/Oct/20

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

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

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

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