求sin(cosA)的极值
求 \(\sin(\cos A)\) 的极值 极大值Sin(1) zeroieme 发表于 2018-1-12 20:02极大值Sin(1)
\(\D1，求证：\lim_{\theta\to o}\ \ \frac{\sin\theta-\tan(\tan\theta)}{\sin(\sin\theta)-\tan\theta}=\frac{5}{4}\)
\(\D2，求证：\lim_{\theta\to o}\ \ \frac{\sin(\sin\theta)-\tan\theta}{\theta\cos\theta-\theta}=\frac{4}{3}\)
\(\D3，求证：\lim_{\theta\to o}\ \ \frac{\tan(\sin\theta)-\tan(\sin(\sin\theta))+\tan(\tan\theta)-\tan\theta}{\sin\theta-\sin(\sin\theta)+\sin(\tan\theta)-\sin(\sin(\tan\theta))}=\frac{3}{2}\)
\(\D4，求证：\lim_{\theta\to o}\ \ \frac{\tan(\tan(\tan(\tan\theta)))-\tan(\tan(\tan\theta))+\tan(\tan\theta)-\tan\theta}{\sin\theta-\sin(\sin\theta)+\sin(\sin(\sin\theta))-\sin(\sin(\sin(\sin\theta)))}=\frac{2}{1}\) 王守恩 发表于 2018-1-13 12:33
\(\D1，求证：\lim_{\theta\to o}\ \ \frac{\sin\theta-\tan(\tan\theta)}{\sin(\sin\theta)-\tan\theta} ...
洛必达法则
https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
https://zh.wikipedia.org/wiki/%E6%B4%9B%E5%BF%85%E8%BE%BE%E6%B3%95%E5%88%99 zeroieme 发表于 2018-1-13 13:45
洛必达法则
https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
https://zh.wikipedia.org/wiki ...
第1个链接中的1道题：(原题的解法)
\(\D\lim_{\theta\to 0}\) \(\frac{2\sin\theta-\sin 2\theta}{\theta-\sin\theta}=\frac{2\cos\theta-2\cos 2\theta}{1-\cos\theta}=\frac{-2\sin\theta+4\sin 2\theta}{\sin\theta}=\frac{-2\cos\theta+8\cos 2\theta}{\cos\theta}=\frac{-2+8}{1}=6\)
下面的解法，请zeroieme检查，有问题吗？
\(\D\lim_{\theta\to 0} \frac{2\sin\theta-\sin 2\theta}{\theta-\sin\theta}=\frac{2×(\theta-\frac{1}{6}\theta^3)-}{\theta-(\theta-\frac{1}{6}\theta^3)}=\frac{2\theta-\frac{2}{6}\theta^3-2\theta+\frac{8}{6}\theta^3}{\frac{1}{6}\theta^3}=\frac{\frac{-2+8}{6}\theta^3}{\frac{1}{6}\theta^3}=6\)
\(\D\begin{align*}\lim_{\theta\to 0} \frac{2 \sin \theta-\sin 2 \theta}{\theta-\sin\theta}&=\frac{2\sin\theta-2\sin\theta\cos\theta}{\theta-\sin\theta}=\frac{2\sin\theta(1-\cos\theta)}{\theta-\sin\theta}=\frac{2×(\theta-\frac{1}{6}\theta^3)}{\theta-(\theta-\frac{1}{6}\theta^3)}\\
&=\frac{2×(\theta-\frac{1}{6}\theta^3)×\frac{1}{2}\theta^2}{\frac{1}{6}\theta^3}=\frac{\theta^3+0(\theta^3)}{\frac{1}{6}\theta^3}=6\end{align*}\) zeroieme 发表于 2018-1-13 13:45
洛必达法则
https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
https://zh.wikipedia.org/wiki ...
下面的解法，省略了过程，很不好意思，麻烦zeroieme，检查核实，有问题吗？
\(\D 1，\lim_{\theta\to 0}\ \ \frac{\sin 2\theta - \tan 2\theta}{\theta-\tan(\tan \theta)}=6\)
\(\D 2，\lim_{\theta\to 0}\ \ \frac{9\sin \theta - \sin 9\theta}{5\sin \theta-\sin 5\theta}=6\)
\(\D 3，\lim_{\theta\to 0}\ \ \frac{\tan 2\theta - 2\tan \theta}{\tan \theta-\theta}=6\)
\(\D 4，\lim_{\theta\to 0}\ \ \frac{3\sin \theta - \sin 3\theta}{\tan \theta-\sin(\sin\theta)}=6\)
\(\D 5，\lim_{\theta\to 0}\ \ \frac{\tan \theta - \sin \theta \cos \theta}{\tan \theta \cos \theta-\sin(\sin \theta)}=6\)
\(\D 6，\lim_{\theta\to 0}\ \ \frac{3\tan \theta\tan \theta - 3\sin \theta\sin \theta}{\sin \theta\tan \theta-\sin \theta\sin \theta}=6\)
本帖最后由 王守恩 于 2018-1-16 14:26 编辑
zeroieme 发表于 2018-1-13 13:45
洛必达法则
https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
https://zh.wikipedia.org/wiki ...
再来几朵小花，希望大家喜欢。
\(\D 1，\lim_{\theta\to 0}\ \ \ \frac{3\sin\theta-\sin3\theta}{2\sin\theta-\sin2\theta}=6\)
\(\D 2，\lim_{\theta\to 0}\ \ \ \frac{3\tan\theta-\tan3\theta}{2\tan\theta-\tan2\theta}=6\)
\(\D 3，\lim_{\theta\to 0}\ \ \ \frac{9\sin\theta-\sin9\theta}{5\sin\theta-\sin5\theta}=6\)
\(\D 4，\lim_{\theta\to 0}\ \ \ \frac{9\tan\theta-\tan9\theta}{5\tan\theta-\tan5\theta}=6\)
\(\D 5，\lim_{\theta\to 0}\ \ \ \frac{9\cos\theta-9\cos9\theta}{5\cos\theta-5\cos5\theta}=6\)
本帖最后由 王守恩 于 2018-1-18 10:33 编辑
zeroieme 发表于 2018-1-13 13:45
洛必达法则
https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
https://zh.wikipedia.org/wiki ...
谢谢知识渊博的zeroieme先生！给我漂亮的链接。
谢谢漂亮的链接，给我灵感：只要4个\(\D\theta\)，可以得到任意自然数！
\(\D\lim_{\theta\to0}\ \frac{\tan\theta-\tan(\sin\theta)}{\sin\theta-\sin(\sin\theta)}=1\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan\theta)-\tan(\theta)}{\sin\theta-\sin(\sin\theta)}=2\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan\theta)-\tan(\sin\theta)}{\sin\theta-\sin(\sin\theta)}=3\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan(tan\theta))-\tan(\theta)}{\sin\theta-\sin(\sin\theta)}=4\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan(\tan\theta))-\tan(\sin\theta)}{\sin\theta-\sin(\sin\theta)}=5\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan(\tan(\tan\theta)))-\tan(\theta)}{\sin\theta-\sin(\sin\theta)}=6\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan(\tan(\tan\theta)))-\tan(\sin\theta)}{\sin\theta-\sin(\sin\theta)}=7\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan(\tan(\tan(tan\theta))))-\tan(\theta)}{\sin\theta-\sin(\sin\theta)}=8\)
\(\D\lim_{\theta\to0}\ \frac{\tan(\tan(\tan(\tan(\tan\theta))))-\tan(\sin\theta)}{\sin\theta-\sin(\sin\theta)}=9\)
................ 王守恩 发表于 2018-1-18 10:24
谢谢知识渊博的zeroieme先生！给我漂亮的链接。
谢谢漂亮的链接，给我灵感：只要4个\(\D\theta\)，可 ...
知识渊博不敢接受，这只是微积分入门课程，离本论坛各位大佬的水平还有几光年。 王守恩 发表于 2018-1-18 10:24
谢谢知识渊博的zeroieme先生！给我漂亮的链接。
谢谢漂亮的链接，给我灵感：只要4个\(\D\theta\)，可 ...
短是短了，但好像有点乱了。
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 7\theta-\cos 5\theta}{\cos 5\theta-\cos \theta}=1\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 5\theta-\cos 3\theta}{\cos 3\theta-\cos \theta}=2\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 4\theta-\cos \theta}{\cos 3\theta-\cos 2\theta}=3\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 6\theta-\cos 4\theta}{\cos 3\theta-\cos 2\theta}=4\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 6\theta-\cos \theta}{\cos 4\theta-\cos 3\theta}=5\)
\(\D\lim_{\theta\to0}\ \ \ \ \ \ \frac{\cos 7\theta-\cos \theta}{\cos 3\theta-\cos \theta}=6\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 8\theta-\cos \theta}{\cos 5\theta-\cos 4\theta}=7\)
\(\D\lim_{\theta\to0}\ \ \ \frac{\cos 11\theta-\cos 7\theta}{\cos 5\theta-\cos 4\theta}=8\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 10\theta-\cos \theta}{\cos 6\theta-\cos 5\theta}=9\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 11\theta-\cos \theta}{\cos 4\theta-\cos 2\theta}=10\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 12\theta-\cos \theta}{\cos 7\theta-\cos 6\theta}=11\)
\(\D\lim_{\theta\to0}\ \ \frac{\cos 16\theta-\cos 10\theta}{\cos 7\theta-\cos 6\theta}=12\)
\(\D\lim_{\theta\to0}\ \ \ \ \ \frac{\cos 14\theta-\cos \theta}{\cos 8\theta-\cos 7\theta}=13\)
\(\D\lim_{\theta\to0}\ \ \ \ \ \frac{\cos 15\theta-\cos \theta}{\cos 5\theta-\cos 3\theta}=14\)
\(\D\lim_{\theta\to0}\ \ \ \ \ \frac{\cos 16\theta-\cos \theta}{\cos 9\theta-\cos 8\theta}=15\)
\(\D\lim_{\theta\to0}\ \ \frac{\cos 21\theta-\cos 13\theta}{\cos 9\theta-\cos 8\theta}=16\)
\(\D\lim_{\theta\to0}\ \ \ \ \frac{\cos 18\theta-\cos \theta}{\cos 10\theta-\cos 9\theta}=17\)
\(\D\lim_{\theta\to0}\ \ \ \ \ \ \frac{\cos 19\theta-\cos \theta}{\cos 6\theta-\cos 4\theta}=18\)
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