Question Number 196419 by Rodier97 last updated on 25/Aug/23
$$\:\:{calcul}\:\:{lim}_{{x}\rightarrow+\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{{f}\left({x}\right)} \\ $$
Commented by MM42 last updated on 24/Aug/23
$${f}\left(\infty\right)\:? \\ $$
Answered by mr W last updated on 24/Aug/23
$${if}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)={a}\neq\mathrm{0}: \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{{f}\left({x}\right)} =\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)^{{a}} \\ $$$$ \\ $$$${if}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0}: \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{{f}\left({x}\right)} {doesn}'{t}\:{exist}. \\ $$$$ \\ $$$${if}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=+\infty: \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{{f}\left({x}\right)} ={e} \\ $$$$ \\ $$$${if}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=−\infty: \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{f}\left({x}\right)}\right)^{{f}\left({x}\right)} =\frac{\mathrm{1}}{{e}} \\ $$