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Question Number 113511 by Khalmohmmad last updated on 13/Sep/20

Answered by mathmax by abdo last updated on 13/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln \left(\mathrm{lnx}\right)}{\mathrm{x}}\mathrm{changement}\mathrm{lnx}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{e}}^{\mathrm{t}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln \left(\mathrm{t}\right)}{{\mathrm{e}}^{\mathrm{t}}}={\mathrm{e}}^{-\mathrm{t}}\ln \left(\mathrm{t}\right)\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{t}\to +\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\ln \left(\mathrm{t}\right)=0$$

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