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Question Number 196419 by Rodier97 last updated on 25/Aug/23

$$calcul{lim}_{x\to +\mathrm{\infty }}{(1+\frac{1}{f\left(x\right)})}^{f\left(x\right)}$$

Commented by MM42 last updated on 24/Aug/23

$$f\left(\mathrm{\infty }\right)?$$

Answered by mr W last updated on 24/Aug/23

$$if\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)=a\ne 0:$$$$\underset{x\to +\mathrm{\infty }}{\lim }{(1+\frac{1}{f\left(x\right)})}^{f\left(x\right)}={(1+\frac{1}{a})}^a$$$$$$$$if\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)=0:$$$$\underset{x\to +\mathrm{\infty }}{\lim }{(1+\frac{1}{f\left(x\right)})}^{f\left(x\right)}{doesn}^{\prime }texist.$$$$$$$$if\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)=+\mathrm{\infty }:$$$$\underset{x\to +\mathrm{\infty }}{\lim }{(1+\frac{1}{f\left(x\right)})}^{f\left(x\right)}=e$$$$$$$$if\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)=-\mathrm{\infty }:$$$$\underset{x\to +\mathrm{\infty }}{\lim }{(1+\frac{1}{f\left(x\right)})}^{f\left(x\right)}=\frac{1}{e}$$

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