Question Number 63959 by meme last updated on 11/Jul/19
$${li}\underset{{x}\rightarrow+\infty} {{m}e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \\ $$
Commented by Prithwish sen last updated on 11/Jul/19
$$\:\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} \rightarrow\:\mathrm{e} \\ $$
Commented by Mikael last updated on 11/Jul/19
$${e}^{\underset{{x}\rightarrow\infty} {{lim}}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \:} =\:{e}^{{ln}\:\underset{{x}\rightarrow\infty} {{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} } =\:{e}^{{lne}} \:=\:{e} \\ $$