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Question Number 113530 by Khalmohmmad last updated on 13/Sep/20

Answered by Aziztisffola last updated on 13/Sep/20

$$-\mathrm{\infty }\times (+\mathrm{\infty })=-\mathrm{\infty }$$

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

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{lnx}}{\mathrm{x}}\mathrm{changement}\mathrm{x}=\frac{1}{\mathrm{t}}\mathrm{give}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{1}{\mathrm{t}}\right)=-\mathrm{tln}\left(\mathrm{t}\right)$$$$\mathrm{x}\to 0^{+}\Rightarrow \mathrm{t}\to +\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{t}\to +\mathrm{\infty }}-\mathrm{tlnt}=-\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{x}\to 0^{+}}\frac{\mathrm{lnx}}{\mathrm{x}}=-\mathrm{\infty }$$

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