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Question Number 144513 by mnjuly1970 last updated on 26/Jun/21

Answered by Olaf_Thorendsen last updated on 26/Jun/21

$${(1+\frac{1}{x})}^{-x}=e^{-x\ln (1+\frac{1}{x})}\underset{\mathrm{\infty }}{\sim }{e}^{-x(\frac{1}{x}-\frac{1}{2x^2})}$$$${(1+\frac{1}{x})}^{-x}\underset{\mathrm{\infty }}{\sim }\frac{1}{e}{e}^{\frac{1}{2x}}$$$$$$$${(1-\frac{1}{x})}^x=e^{x\ln (1-\frac{1}{x})}\underset{\mathrm{\infty }}{\sim }{e}^{x(-\frac{1}{x}-\frac{1}{2x^2})}$$$${(1-\frac{1}{x})}^x\underset{\mathrm{\infty }}{\sim }\frac{1}{e}{e}^{-\frac{1}{2x}}$$$$$$$$f\left(x\right)=x[{(1+\frac{1}{x})}^{-x}-{(1-\frac{1}{x})}^x]$$$$f\left(x\right)\underset{\mathrm{\infty }}{\sim }\frac{x}{e}[e^{\frac{1}{2x}}-e^{-\frac{1}{2x}}]$$$$f\left(x\right)\underset{\mathrm{\infty }}{\sim }\frac{2x.\sinh \left(\frac{1}{2x}\right)}{e}\underset{\mathrm{\infty }}{\sim }\frac{2x.\frac{1}{2x}}{e}=\frac{1}{e}$$

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