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[函数] $(1+x)^{1/x}-x^{1/x}$渐近展开

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hbghlyj Posted at 2023-3-23 22:20:42 |Read mode
Last edited by hbghlyj at 2023-5-30 15:13:00当$x>1$时$$(1+x)^{x^{-1}}-x^{x^{-1}}>x^{-2}$$
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realnumber Posted at 2023-5-31 07:24:53
$y=x^{\frac{1}{x}}((1+\frac{1}{x})^\frac{1}{x}-1)>x^\frac{1}{x}(1+\frac{1}{x^2}-1)=x^\frac{1}{x}\frac{1}{x^2}>\frac{1}{x^2}$


提个公因式后,用了下伯努利不等式-对任意实数n≥0,和任意实数x≥-1,有 (1+x)^n≥1+nx

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