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本帖最后由 王守恩 于 2020-11-23 08:19 编辑
elim 发表于 2020-11-23 02:13
例 试证 \(\beta=\frac{1+\sqrt{5}}{2},\;\,\displaystyle\lim_{n\to\infty}\big(n^{\beta}-(n-\frac{\pi}{ ...
例 试证 \(\displaystyle\beta=\frac{1+\sqrt{5}}{2},\ \lim_{n\to\infty}\big(n^{\beta}-(n-\frac{\pi}{\beta\ n^{1/\beta}})^{\beta}\big)=\pi\)
\(我们有:\displaystyle\lim_{x\to\infty}\big( x^a-(\frac{y}{ax^{a-1}}+x)^a\big)=-y\ \ 可得\)
\(\displaystyle\ \ \lim_{x\to\infty}\big(x^a-(x-\frac{y}{ax^{a-1}})^a\big)=y\ \ \ \ (1)\)
\(\beta=\frac{1+\sqrt{5}}{2},\ 则\frac{1}{\beta}=\frac{\sqrt{5}-1}{2}\)
\(\beta=\frac{1+\sqrt{5}}{2},\ 则\beta-1=\frac{\sqrt{5}-1}{2}\)
\(即:\frac{1}{\beta}=\frac{\sqrt{5}-1}{2}=\beta-1\)
\(\displaystyle\lim_{n\to\infty}\big(n^{\beta}-(n-\frac{\pi}{\beta n^{1/\beta}})^{\beta}\big)\)
\(\displaystyle=\lim_{n\to\infty}\big(n^{\beta}-(n-\frac{\pi}{\beta n^{\beta-1}})^{\beta}\big)\ \ \frac{(1)}{=}\ \ \pi\) |
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