证明0.999...=1 《高等数学》复旦版黄立宏第四版 Posted on 2021-01-29 11:30:00 In Mathematics 根据数列极限的定义证明：limn→∞0.99⋯9⏞n个=1.\lim \limits_{n \to \infty} 0.\overset{n个}{\overbrace{ 99 \cdots 9 } } = 1.n→∞lim0.99⋯9n个=1. 证明：因为对于所有的正整数 n ，有∣0.999⋯9⏞n个−1∣<1\left | 0.\overset{n个}{\overbrace{999 \cdots 9}} - 1 \right | < 1∣∣∣∣∣0.999⋯9n个−1∣∣∣∣∣<1，故 ∀ ε > 0，不妨设 ε < 1，要使 ∣0.99⋯9⏞n个−1∣=110n<ε\left | 0.\overset{n个}{\overbrace{99 \cdots 9}} - 1 \right | = \frac{1}{10^{n}} < \varepsilon∣∣∣∣∣0.99⋯9n个−1∣∣∣∣∣=10n1<ε ，只要 n>−lnεln10n > \frac{-\ln \varepsilon }{\ln 10}n>ln10−lnε，取N=[−lnεln10]N = \left [ \frac{-\ln \varepsilon }{\ln 10} \right ]N=[ln10−lnε] ，则当 n > N 时，恒有 ∣0.99⋯9⏞n个−1∣<ε\left | 0.\overset{n个}{\overbrace{99 \cdots 9}} - 1 \right | < \varepsilon∣∣∣∣∣0.99⋯9n个−1∣∣∣∣∣<ε，故 limn→∞0.99⋯9⏞n个=1.\lim \limits_{n \to \infty} 0.\overset{n个}{\overbrace{ 99 \cdots 9 } } = 1.n→∞lim0.99⋯9n个=1. Buy me a coffee Donate Alipay
