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

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\left(\frac{{\mathrm{n}}^3+1}{{\mathrm{n}}^3-1}\right)}^{2\mathrm{n}-{\mathrm{n}}^3}$$

Commented by Tawa1 last updated on 01/Jul/19

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

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

$$let{A}_n={\left(\frac{n^3+1}{n^3-1}\right)}^{2n-n^3}\Rightarrow A_n={(1-\frac{2}{n^3-1})}^{2n-n^3}=e^{(2n-n^3)ln(1-\frac{2}{n^3-1})}wehave$$$$ln(1-x)=-x+o\left(x^2\right)\Rightarrow ln(1-x)\sim -x(x\to 0)\Rightarrow ln(1-\frac{2}{n^3-1})\sim -\frac{2}{n^3-1}(n\to +\mathrm{\infty })$$$$(2n-n^3)ln(1-\frac{2}{n^3-1})\sim \frac{-2(2n-n^3)}{n^3-1}=\frac{2n^3-4n}{n^3-1}\sim 2(n\to +\mathrm{\infty })\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{A}_n=e^2.$$

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

$$youarewelcome.$$

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