计算 \(\displaystyle\lim_{x\to 0}\sqrt[\large x^2]{\cos x}\)

luyuanhong

楼上 elim 的解答很好！已收藏。

王守恩

王守恩 发表于 2021-12-28 07:03
我只是凑凑热闹(内心深处还是想讨教)。

\(\displaystyle\lim_{x\to 0}\ \sqrt[x^2]{\cos\big(\frac{ax ...

\((1)与(2)是互为倒数关系。\)

\(\displaystyle\lim_{x\to 0}\ \sqrt[x^2]{\cos(\sqrt{2}x)}=\frac{1}{e}\ \ \ \ \ \ (1)\)

\(\displaystyle\lim_{x\to 0}\sqrt[x^2]{\cos(\sqrt{2}ix)}=e=\lim_{x\to 0}\sqrt[x^2]{\cosh(\sqrt{2}x)}\ \ \ \ (2)\)

elim

\(\displaystyle\lim_{x\to 0} \sqrt[\Large x^2]{\cos(\lambda x)} = \exp\left(\lim_{x\to 0}x^{-2}\ln(1-(1-\cos(\lambda x)))\right)\)
\(\quad =\displaystyle\exp\left(-\lambda^2\lim_{x\to 0}\frac{1-\cos(\lambda x)}{(\lambda x)^2}\right)=\exp\big(-\frac{\lambda^2}{2}\big)\)
\(\displaystyle\lim_{x\to 0} \sqrt[\Large x^2]{\cos(\sqrt{2} x)} =\exp\big(-\frac{(\sqrt{2})^2}{2}\big)=\frac{1}{e}\)
\(\displaystyle\lim_{x\to 0} \sqrt[\Large x^2]{\cos(\sqrt{2}i x)} =\exp\big(\frac{(\sqrt{2})^2}{2}\big)=e\)

王守恩

elim 发表于 2021-12-29 04:08
\(\displaystyle\lim_{x\to 0} \sqrt[\Large x^2]{\cos(\lambda x)} = \exp\left(\lim_{x\to 0}x^{-2}\ln(1 ...

谢谢 elim！收获不少，画蛇添足还是来一下。 谢谢 elim！

\(\displaystyle\lim_{x\to 0}\sqrt[x^2]{\cos\big(\lambda\ x\big)}=\lim_{x\to 0}\sqrt[x^2]{\cosh\big(\lambda\ i\ x\big)}=\lim_{x\to\infty}\big(\cos\frac{\lambda}{x}\big)^{x^2}=\lim_{x\to\infty}\big(\cosh\frac{\lambda\ i}{x}\big)^{x^2}=\sum_{k=0}^{\infty}\frac{(\frac{\lambda^2}{-2})^k}{k!}=\exp\big(\frac{\lambda^2}{-2}\big)\)

\(\displaystyle\lim_{x\to 0}\sqrt[x^2]{\cos\big(\lambda\ i\ x\big)}=\lim_{x\to 0}\sqrt[x^2]{\cosh\big(\lambda\ x\big)}=\lim_{x\to\infty}\big(\cos\frac{\lambda\ i}{x}\big)^{x^2}=\lim_{x\to\infty}\big(\cosh\frac{\lambda}{x}\big)^{x^2}=\sum_{k=0}^{\infty}\frac{(\frac{\lambda^2}{2})^k}{k!}=\exp\big(\frac{\lambda^2}{2}\big)\)

王守恩

王守恩 发表于 2021-12-25 12:06
我只是凑凑热闹。

\(\displaystyle\lim_{x\to 0}\sqrt[\large x^2]{\cos x}=\lim_{x\to\infty}\cos^{x^2 ...

我只是凑凑热闹。谢谢 elim！
有这样一串数：可是在《整数序列在线百科全书(OEIS)》找不到的。

{1, 2, 3, 4, 7, 12, 20, 33, 55, 90, 148, 245, 403, 665, 1097, 1808, 2981, 4915, 8103,
  13360, 22026, 36316, 59874, 98716, 162755, 268337, 442413, 729416, 1202604,
  1982759, 3269017, 5389698, 8886111, 14650719, 24154953, 39824784, .......}

通项公式是这样：\(\lambda=0, 1, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, .....\)
\(\displaystyle\lim_{x\to 0}[\sqrt[x^2]{\cos(\lambda ix)}\ ]=\lim_{x\to 0}[\sqrt[x^2]{\cosh(\lambda x)}\ ]=\lim_{x\to\infty}[(\cos\frac{\lambda i}{x})^{x^2}]=\lim_{x\to\infty}[(\cosh\frac{\lambda}{x})^{x^2}]=[\sum_{k=0}^{\infty}\frac{(\frac{\lambda^2}{2})^k}{k!}]=\exp[\frac{\lambda^2}{2}]\)

elim

请王守恩老师分享一下为什么 \(\displaystyle\lim_{x\to\infty}\left(\cosh\frac{\lambda}{x}\right)^{x^2}=\exp\big(\frac{\lambda^2}{2}\big)\) ？

王守恩

elim 发表于 2021-12-29 12:47
请王守恩老师分享一下为什么 \(\displaystyle\lim_{x\to\infty}\left(\cosh\frac{\lambda}{x}\right)^{x^2} ...

只能回归13楼。还望 elim 赐教。

\(\displaystyle\cos(\frac{\lambda}{x})=\frac{(\frac{\lambda}{x})^0}{0!}-\frac{(\frac{\lambda}{x})^2}{2!}+\frac{(\frac{\lambda}{x})^4}{4!}-\frac{(\frac{\lambda}{x})^6}{6!}+\frac{(\frac{\lambda}{x})^8}{8!}-...\)

\(\displaystyle\cosh(\frac{\lambda\ i}{x})=\frac{(\frac{\lambda}{x})^0}{0!}-\frac{(\frac{\lambda}{x})^2}{2!}+\frac{(\frac{\lambda}{x})^4}{4!}-\frac{(\frac{\lambda}{x})^6}{6!}+\frac{(\frac{\lambda}{x})^8}{8!}-...\)

\(\displaystyle\exp(\frac{\lambda}{-2})=\frac{(\frac{\lambda}{2})^0}{0!}-\frac{(\frac{\lambda}{2})^1}{1!}+\frac{(\frac{\lambda}{2})^2}{2!}-\frac{(\frac{\lambda}{2})^3}{3!}+\frac{(\frac{\lambda}{2})^4}{4!}-...\)

\(\displaystyle\cos(\frac{\lambda\ i}{x})=\frac{(\frac{\lambda}{x})^0}{0!}+\frac{(\frac{\lambda}{x})^2}{2!}+\frac{(\frac{\lambda}{x})^4}{4!}+\frac{(\frac{\lambda}{x})^6}{6!}+\frac{(\frac{\lambda}{x})^8}{8!}+...\)

\(\displaystyle\cosh(\frac{\lambda}{x})=\frac{(\frac{\lambda}{x})^0}{0!}+\frac{(\frac{\lambda}{x})^2}{2!}+\frac{(\frac{\lambda}{x})^4}{4!}+\frac{(\frac{\lambda}{x})^6}{6!}+\frac{(\frac{\lambda}{x})^8}{8!}+...\)

\(\displaystyle\exp(\frac{\lambda}{2})=\frac{(\frac{\lambda}{2})^0}{0!}+\frac{(\frac{\lambda}{2})^1}{1!}+\frac{(\frac{\lambda}{2})^2}{2!}+\frac{(\frac{\lambda}{2})^3}{3!}+\frac{(\frac{\lambda}{2})^4}{4!}+...\)

elim

楼上的这些式子没有问题，只是\(\cosh^{x^2}\big({\large\frac{\lambda}{x}}\big)\) 的极限怎么算，还是得回到我的证法那里去.
下面几个结果是被严格论证和人们熟知的：
\((1)\quad \lim f(x)^{g(x)}=\exp(\lim g(x)\ln f(x))\)
\((2) \quad \ln(1+u) = u+O(u^2)\quad (u\to 0)\)
\((3)\quad \displaystyle\lim_{t\to 0}\frac{1-\cos t}{t^2}=\frac{1}{2}\)

王守恩

王守恩 发表于 2021-12-29 12:21
我只是凑凑热闹。谢谢 elim！
有这样一串数：可是在《整数序列在线百科全书(OEIS)》找不到的。

我只是凑凑热闹。谢谢 elim！

{1, 2, 3, 4, 7, 12, 20, 33, 55, 90, 148, 245, 403, 665, 1097, 1808, 2981, 4915, 8103,
  13360, 22026, 36316, 59874, 98716, 162755, 268337, 442413, 729416, 1202604,
  1982759, 3269017, 5389698, 8886111, 14650719, 24154953, 39824784, .......}

\(通项公式=[\exp(\frac{\lambda}{2})]\ \ \ \lambda=0, 1, 2, 3, 4, .....\)

她的倒数和好像是这样

\(\displaystyle\sum_{\lambda=0}^{\infty}\exp\big(\frac{-\lambda}{2}\big)=\frac{\sqrt{e}}{\sqrt{e}-1}\)

王守恩

elim 发表于 2021-12-30 11:02
楼上的这些式子没有问题，只是\(\cosh^{x^2}\big({\large\frac{\lambda}{x}}\big)\) 的极限怎么算，还是得 ...

这也是一串不错的数(慢慢在捂)。

\(n=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29\)

\(\displaystyle\lim_{t\to 0}\frac{\sin(t\ n)}{t}=a(n)\)

\(a(n)=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29\)