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Question Number 106662 by mathdave last updated on 06/Aug/20

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

$$\mathrm{Let}{\mathrm{A}}_{\mathrm{n}}=\underset{\mathrm{x}\to 0}{\lim }\ln {(\mathrm{x}!)}^{\frac{1}{\mathrm{x}}},\mathrm{x}\in \mathbb{N}$$$$\mathrm{lnx}!=\ln (\left(\mathrm{x}\right)(\mathrm{x}-1)(\mathrm{x}-2)...(\mathrm{x}-(\mathrm{x}-1)))$$$$=\ln \underset{\mathrm{k}=0}{\prod ^{\mathrm{x}-1}}(\mathrm{x}-\mathrm{k})=\underset{\mathrm{k}=0}{\sum ^{\mathrm{x}-1}}\ln (\mathrm{x}-\mathrm{k})$$$${\mathrm{A}}_{\mathrm{n}}=\underset{\mathrm{x}\to 0}{\lim }\left\{\frac{1}{\mathrm{x}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{x}-1}}\ln (\mathrm{x}-\mathrm{k})\right\}?????$$$$$$$$\mathrm{However},\mathrm{if}\mathrm{x}\to \mathrm{\infty }{\mathrm{we}}^{\prime }\mathrm{ll}\mathrm{have};$$$${\mathrm{A}}_{\mathrm{n}}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\left\{\frac{1}{\mathrm{x}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{x}-1}}\ln (\mathrm{x}-\mathrm{k})\right\}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{\int }_0^1\mathrm{lnx}(1-\mathrm{t})\mathrm{dt}$$$$=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{\int }_0^1\ln \left(\mathrm{ux}\right)\mathrm{du}=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{\left\{\frac{\mathrm{ux}}{\mathrm{x}}(\ln \left(\mathrm{ux}\right)-1)\right\}}_0^1$$$$=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\{(\mathrm{lnx}-1)-\underset{\mathrm{u}\to 0}{\lim }[\mathrm{uln}\left(\mathrm{ux}\right)-\mathrm{u}]\}$$$$=\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\{\mathrm{lnx}-1\}=+\mathrm{\infty }$$

Commented by mathdave last updated on 07/Aug/20

$$nodansweris=1$$$$$$

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