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Question Number 175994 by sciencestudent last updated on 10/Sep/22

$$li\underset{x\to \mathrm{\infty }}{m}\frac{e^x-1}{x}=?$$

Answered by JDamian last updated on 10/Sep/22

$$\mathrm{\infty }$$

Answered by cortano2 last updated on 11/Sep/22

$$=\frac{e^{\mathrm{\infty }}-1}{\mathrm{\infty }}=\frac{\mathrm{\infty }}{\mathrm{\infty }}=1$$

Commented by Ar Brandon last updated on 11/Sep/22

$$e^x\gg x\mathrm{as}x\to +\mathrm{\infty }$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^x-1}{x}\to +\mathrm{\infty }$$

Answered by Raxreedoroid last updated on 11/Sep/22

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{e^x-1}{x}\left(\frac{\mathrm{\infty }}{\mathrm{\infty }}\right)$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{e^x}{1}=+\mathrm{\infty }$$

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