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Question Number 183693 by LEKOUMA last updated on 28/Dec/22

$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{{\left(\ln x\right)}^x}{x^{\ln x}}$$

Answered by a.lgnaoui last updated on 28/Dec/22

$$\mathrm{lnx}=\mathrm{t}\Rightarrow \mathrm{x}={\mathrm{e}}^{\mathrm{t}}$$$$\mathrm{x}\to +\mathrm{\infty }\Rightarrow t\to +\mathrm{\infty }$$$$\frac{t^{e^t}}{{\left(e^t\right)}^t}=\frac{t^{e^t}}{e^{t^2}}\Rightarrow \ln \left(\frac{{\mathrm{t}}^{{\mathrm{e}}^{\mathrm{t}}}}{{\mathrm{e}}^{{\mathrm{t}}^2}}\right)={\mathrm{e}}^{\mathrm{t}}\ln \left(\mathrm{t}\right)-{\mathrm{t}}^2$$$$\mathrm{t}\Rightarrow +\mathrm{\infty }{e}^t(\ln \left(\mathrm{t}\right)-\frac{t^2}{e^t})\Rightarrow +\mathrm{\infty }$$$$donc{\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{{\left(\mathrm{lnx}\right)}^{\mathrm{x}}}{{\mathrm{x}}^{\mathrm{lnx}}}=+\mathrm{\infty }$$

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