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Question Number 98833 by john santu last updated on 16/Jun/20

Commented by bobhans last updated on 16/Jun/20

lim_(n→∞) (3^n (1+((1/6^n )))^(1/n) ) = 3 lim_(n→∞) (1+(1/6^n ))^(1/n) = 3 ×1 = 3

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{3}^{\mathrm{n}} \:\left(\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \right)\:=\:\mathrm{3}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$$$=\:\mathrm{3}\:×\mathrm{1}\:=\:\mathrm{3}\: \\ $$

Answered by mathmax by abdo last updated on 16/Jun/20

let A_n =(3^n +(1/2^n ))^(1/n) ⇒ A_n =3(1+(1/6^n ))^n =3 e^(nln(1+(1/6^n ))) but ln(1+(1/6^n )) ∼(1/6^n ) ⇒n on(1+(1/6^n ))∼(n/6^n ) →0(n→+∞) ⇒lim_(n→+∞) A_n =3

$$\mathrm{let}\:\mathrm{A}_{\mathrm{n}} =\left(\mathrm{3}^{\mathrm{n}} \:+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\Rightarrow\:\mathrm{A}_{\mathrm{n}} =\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)^{\mathrm{n}} \:=\mathrm{3}\:\mathrm{e}^{\mathrm{nln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)} \:\:\:\mathrm{but} \\ $$$$\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)\:\sim\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\:\Rightarrow\mathrm{n}\:\mathrm{on}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{n}} }\right)\sim\frac{\mathrm{n}}{\mathrm{6}^{\mathrm{n}} }\:\rightarrow\mathrm{0}\left(\mathrm{n}\rightarrow+\infty\right)\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{A}_{\mathrm{n}} =\mathrm{3} \\ $$

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