Answer

问题及解答

证明: 当 $x > 0$ 时, $e^{\frac{x}{x+1}} < (1+\frac{1}{x})^x < e$.

Posted by haifeng on 2022-11-10 13:21:28 last update 2022-11-10 13:21:28 | Edit | Answers (1)

证明: 当 $x > 0$ 时,

\[e^{\frac{x}{x+1}} < (1+\frac{1}{x})^x < e.\]

1

Posted by haifeng on 2023-11-02 17:37:43

取对数后不等式等价于
\[
\frac{x}{1+x}<x\ln(1+\frac{1}{x})<1.
\]
这等价于
\[
\frac{\frac{1}{x}}{1+\frac{1}{x}}=\frac{1}{1+x}<\ln(1+\frac{1}{x})<\frac{1}{x}.
\]