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Question Number 144307 by mohammad17 last updated on 24/Jun/21

lim_(n→∞) (1+(1/n))^p

$${lim}_{{n}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{p}} \\ $$

Commented by mohammad17 last updated on 24/Jun/21

help me sir please

$${help}\:{me}\:{sir}\:{please} \\ $$

Answered by Dwaipayan Shikari last updated on 24/Jun/21

1

$$\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 25/Jun/21

U_n =(1+(1/n))^p  ⇒U_n =e^(plog(1+(1/n)))  ∼e^(p/n)  →1(n→∞) pour p fixe  ⇒lim_(n→+∞) U_n =1

$$\mathrm{U}_{\mathrm{n}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{p}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{e}^{\mathrm{plog}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)} \:\sim\mathrm{e}^{\frac{\mathrm{p}}{\mathrm{n}}} \:\rightarrow\mathrm{1}\left(\mathrm{n}\rightarrow\infty\right)\:\mathrm{pour}\:\mathrm{p}\:\mathrm{fixe} \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{U}_{\mathrm{n}} =\mathrm{1} \\ $$

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