Question Number 144513 by mnjuly1970 last updated on 26/Jun/21
Answered by Olaf_Thorendsen last updated on 26/Jun/21
![(1+(1/x))^(−x) = e^(−xln(1+(1/x))) ∼_∞ e^(−x((1/x)−(1/(2x^2 )))) (1+(1/x))^(−x) ∼_∞ (1/e)e^(1/(2x)) (1−(1/x))^x = e^(xln(1−(1/x))) ∼_∞ e^(x(−(1/x)−(1/(2x^2 )))) (1−(1/x))^x ∼_∞ (1/e)e^(−(1/(2x))) f(x) = x[(1+(1/x))^(−x) −(1−(1/x))^x ] f(x) ∼_∞ (x/e)[e^(1/(2x)) −e^(−(1/(2x))) ] f(x) ∼_∞ ((2x.sinh((1/(2x))))/e) ∼_∞ ((2x.(1/(2x)))/e) = (1/e)](https://www.tinkutara.com/question/Q144591.png)
$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{−{x}} \:=\:{e}^{−{x}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \:\underset{\infty} {\sim}\:{e}^{−{x}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right)} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{−{x}} \:\underset{\infty} {\sim}\:\frac{\mathrm{1}}{{e}}{e}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$ \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{{x}} \:=\:{e}^{{x}\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)} \:\underset{\infty} {\sim}\:{e}^{{x}\left(−\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right)} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{{x}} \:\underset{\infty} {\sim}\:\frac{\mathrm{1}}{{e}}{e}^{−\frac{\mathrm{1}}{\mathrm{2}{x}}} \\ $$$$ \\ $$$${f}\left({x}\right)\:=\:{x}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{−{x}} −\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{{x}} \right] \\ $$$${f}\left({x}\right)\:\underset{\infty} {\sim}\:\frac{{x}}{{e}}\left[{e}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} −{e}^{−\frac{\mathrm{1}}{\mathrm{2}{x}}} \right] \\ $$$${f}\left({x}\right)\:\underset{\infty} {\sim}\:\frac{\mathrm{2}{x}.\mathrm{sinh}\left(\frac{\mathrm{1}}{\mathrm{2}{x}}\right)}{{e}}\:\underset{\infty} {\sim}\:\frac{\mathrm{2}{x}.\frac{\mathrm{1}}{\mathrm{2}{x}}}{{e}}\:=\:\frac{\mathrm{1}}{{e}} \\ $$