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Question Number 203701 by Noorzai last updated on 26/Jan/24

Answered by mr W last updated on 26/Jan/24

$$0<\frac{2^x}{x!}=\frac{2\times 2\times 2\times 2\times ....\times 2}{1\times 2\times 3\times 4\times ...\times x}<{\left(\frac{2}{3}\right)}^x$$$$0<\underset{x\to \mathrm{\infty }}{\lim }\frac{2^x}{x!}<\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{2}{3}\right)}^x=0$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\frac{2^x}{x!}=0$$

Answered by Mathspace last updated on 30/Jan/24

$$x!\sim x^x{e}^{-x}\sqrt{2\pi x}\left(stirling\right)$$$$and\frac{2^x}{x!}=\frac{e^{xln2}}{x^x{e}^{-x}\sqrt{2\pi x}}$$$$=\frac{x^{-x}{e}^{x+xln2}}{\sqrt{2\pi x}}=\frac{e^{-xlnx+x+xln2}}{\sqrt{2\pi x}}$$$$=\frac{e^{-x(lnx-1-ln2)}}{\sqrt{2\pi x}}\sim \frac{e^{-xlnx}}{\sqrt{2\pi x}}\to 0(x\to +\mathrm{\infty })$$$$\Rightarrow {lim}_{x\to +\mathrm{\infty }}\frac{2^x}{x!}=0$$

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