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Question Number 132138 by benjo_mathlover last updated on 11/Feb/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }(\frac{{(1+\frac{1}{\mathrm{n}})}^{\mathrm{n}}}{{(1-\frac{1}{\mathrm{n}})}^{\mathrm{n}}}-{\mathrm{e}}^2){\mathrm{n}}^2=?$$

Answered by EDWIN88 last updated on 11/Feb/21

$$\mathrm{L}={\mathrm{e}}^2\times \underset{x\to 0}{\lim }\frac{\ln (1+\mathrm{x})-\ln (1-\mathrm{x})-2\mathrm{x}}{{\mathrm{x}}^3}$$$$\mathrm{L}={\mathrm{e}}^2\times \underset{x\to 0}{\lim }\frac{\frac{1}{1+\mathrm{x}}+\frac{1}{1-\mathrm{x}}-2}{{3\mathrm{x}}^2}$$$$\mathrm{L}={\mathrm{e}}^2\times \underset{x\to 0}{\lim }\frac{2-2(1-{\mathrm{x}}^2)}{(1+\mathrm{x})(1-\mathrm{x}){3\mathrm{x}}^2}$$$$\mathrm{L}={\mathrm{e}}^2\times \underset{x\to 0}{\lim }\frac{{2\mathrm{x}}^2}{{3\mathrm{x}}^2}=\frac{{2\mathrm{e}}^2}{3}$$

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