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Question Number 143012 by Study last updated on 08/Jun/21

$$li\underset{\alpha \to \mathrm{\infty }}{m}\frac{e^x}{{\alpha }^{60!}}=?$$

Commented by Study last updated on 08/Jun/21

$$isthe{e}^{x}number?$$

Answered by mathmax by abdo last updated on 09/Jun/21

$$\mathrm{for}\mathrm{x}\mathrm{fixed}{\lim }_{\alpha \to \mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{x}}}{{\alpha }^{60!}}=0$$$$\mathrm{perhaps}\mathrm{the}\mathrm{Q}\mathrm{is}{\lim }_{\mathrm{x}\to +\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{x}}^{60!}}$$

Commented by Study last updated on 09/Jun/21

$$whatisthepractice?$$

Commented by mathmax by abdo last updated on 09/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{x}}^{60!}}\Rightarrow \log \left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}-60!\mathrm{logx}=\mathrm{x}(1-60!\frac{\mathrm{logx}}{\mathrm{x}})\Rightarrow$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\log \left(\mathrm{f}\left(\mathrm{x}\right)\right)=+\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=+\mathrm{\infty }$$

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