Question Number 31524 by abdo imad last updated on 09/Mar/18
$${find}\:{lim}_{{n}\rightarrow\infty} \left(\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} . \\ $$
Commented by abdo imad last updated on 12/Mar/18
$${let}\:{put}\:{A}_{{n}} \:=\left(\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:{we}\:{have}\: \\ $$$${ln}\left({A}_{{n}} \right)={nln}\left(\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{{n}}\right)\right)\:\:{but}\:{sin}\left(\frac{\mathrm{1}}{{n}}\right)=\frac{\mathrm{1}}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right){and} \\ $$$${ln}\left(\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{{n}}\right)\right)\:=\:\frac{\mathrm{1}}{{n}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow\:{nln}\left(\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{{n}}\right)\right)=\mathrm{1}+{o}\left(\mathrm{1}\right) \\ $$$${so}\:{lim}_{{n}\rightarrow\infty} {ln}\left({A}_{{n}} \right)=\:\mathrm{1}\:\Rightarrow\:{lim}_{{n}\rightarrow\infty} {A}_{{n}} =\:{e}\:. \\ $$