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Question Number 190679 by mathlove last updated on 09/Apr/23

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{e^x}{x^{60!}}=?$$$$pleassolvethis$$

Answered by Frix last updated on 09/Apr/23

$$\frac{{\mathrm{e}}^x}{x^n}={\mathrm{e}}^{x-n\ln x}$$$$\mathrm{Obviously}\mathrm{for}n\in \mathbb{N}\mathrm{and}x\to +\mathrm{\infty }\mathrm{the}\mathrm{term}$$$$x-n\ln x\to +\mathrm{\infty }\Rightarrow$$$$\mathrm{Answer}\mathrm{is}+\mathrm{\infty }$$

Answered by aba last updated on 09/Apr/23

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{x}}^{60!}}={\underset{x\to \mathrm{\infty }}{\lim e}}^{\mathrm{x}-60!\ln \left(\mathrm{x}\right)}={\underset{x\to \mathrm{\infty }}{\lim e}}^{\mathrm{x}(1-60!\frac{\ln \left(\mathrm{x}\right)}{\mathrm{x}})}=\{\begin{array}{l}0\mathrm{if}\mathrm{x}\to -\mathrm{\infty }\\ +\mathrm{\infty }\mathrm{if}\mathrm{x}\to +\mathrm{\infty }\end{array}$$

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