Question Number 144307 by mohammad17 last updated on 24/Jun/21
$${lim}_{{n}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{p}} \\ $$
Commented by mohammad17 last updated on 24/Jun/21
$${help}\:{me}\:{sir}\:{please} \\ $$
Answered by Dwaipayan Shikari last updated on 24/Jun/21
$$\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 25/Jun/21
$$\mathrm{U}_{\mathrm{n}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{p}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\mathrm{plog}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)} \:\sim\mathrm{e}^{\frac{\mathrm{p}}{\mathrm{n}}} \:\rightarrow\mathrm{1}\left(\mathrm{n}\rightarrow\infty\right)\:\mathrm{pour}\:\mathrm{p}\:\mathrm{fixe} \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} =\mathrm{1} \\ $$