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Question Number 63267 by Tawa1 last updated on 01/Jul/19

lim_(n→∞) (((n^3 + 1)/(n^3 − 1)))^(2n − n^3 )

$$\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\left(\frac{\mathrm{n}^{\mathrm{3}} \:+\:\mathrm{1}}{\mathrm{n}^{\mathrm{3}} \:−\:\mathrm{1}}\right)^{\mathrm{2n}\:−\:\mathrm{n}^{\mathrm{3}} } \\ $$

Commented by Tawa1 last updated on 01/Jul/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mathmax by abdo last updated on 01/Jul/19

let A_n =(((n^3 +1)/(n^3 −1)))^(2n−n^3 ) ⇒ A_n =(1−(2/(n^3 −1)))^(2n−n^3 ) =e^((2n−n^3 )ln(1−(2/(n^3 −1)))) we have ln(1−x) =−x +o(x^2 ) ⇒ln(1−x)∼−x (x→0) ⇒ln(1−(2/(n^3 −1)))∼−(2/(n^3 −1))(n→+∞) (2n−n^3 )ln(1−(2/(n^3 −1))) ∼((−2(2n−n^3 ))/(n^3 −1)) =((2n^3 −4n)/(n^3 −1)) ∼ 2 (n→+∞) ⇒ lim_(n→+∞) A_n =e^2 .

$${let}\:{A}_{{n}} =\left(\frac{{n}^{\mathrm{3}} \:+\mathrm{1}}{{n}^{\mathrm{3}} −\mathrm{1}}\right)^{\mathrm{2}{n}−{n}^{\mathrm{3}} } \:\Rightarrow\:{A}_{{n}} =\left(\mathrm{1}−\frac{\mathrm{2}}{{n}^{\mathrm{3}} −\mathrm{1}}\right)^{\mathrm{2}{n}−{n}^{\mathrm{3}} } ={e}^{\left(\mathrm{2}{n}−{n}^{\mathrm{3}} \right){ln}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}^{\mathrm{3}} −\mathrm{1}}\right)} \:\:{we}\:{have} \\ $$$${ln}\left(\mathrm{1}−{x}\right)\:=−{x}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{ln}\left(\mathrm{1}−{x}\right)\sim−{x}\:\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{ln}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}^{\mathrm{3}} −\mathrm{1}}\right)\sim−\frac{\mathrm{2}}{{n}^{\mathrm{3}} −\mathrm{1}}\left({n}\rightarrow+\infty\right) \\ $$$$\left(\mathrm{2}{n}−{n}^{\mathrm{3}} \right){ln}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}^{\mathrm{3}} −\mathrm{1}}\right)\:\sim\frac{−\mathrm{2}\left(\mathrm{2}{n}−{n}^{\mathrm{3}} \right)}{{n}^{\mathrm{3}} −\mathrm{1}}\:\:=\frac{\mathrm{2}{n}^{\mathrm{3}} −\mathrm{4}{n}}{{n}^{\mathrm{3}} −\mathrm{1}}\:\sim\:\mathrm{2}\:\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ={e}^{\mathrm{2}} . \\ $$

Commented by mathmax by abdo last updated on 02/Jul/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

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