устанавливает фундаментальный предел

\((x - \sin x)/x^{3}\)

НАдо доказать, что формула

\(\phi(x) = x\sin(\sin x) - \sin^{2}x\)

имеет шестой порядок малости, когда \(x\) мал, и найти предел 

"\(\phi(x)/x^{6}\) as \(x \to 0\)" 

Результат:

Используем

\[ \begin{align} \frac{\color{#C00000}{\sin(2x)-2\sin(x)}}{\color{#00A000}{\tan(2x)-2\tan(x)}} &=\underbrace{\color{#C00000}{2\sin(x)(\cos(x)-1)}\vphantom{\frac{\tan^2(x)}{\tan^2(x)}}}\underbrace{\frac{\color{#00A000}{1-\tan^2(x)}}{\color{#00A000}{2\tan^3(x)}}}\\ &=\hphantom{\sin}\frac{-2\sin^3(x)}{\cos(x)+1}\hphantom{\sin}\frac{\cos(x)\cos(2x)}{2\sin^3(x)}\\ &=-\frac{\cos(x)\cos(2x)}{\cos(x)+1}\tag{1} \end{align} \]


Следовательно,

\[ \lim_{x\to0}\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}=-\frac12\tag{2} \]


Таким образом
\(\epsilon\gt0\), найдем \(\delta\gt0\) so that if \(|x|\le\delta\) \[ \left|\,\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}+\frac12\,\right|\le\epsilon\tag{3} \] Because \(\,\displaystyle\lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\frac{\tan(x)}{x}=1\), we have \[ \sin(x)-x=\sum_{k=0}^\infty2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\tag{4} \] and \[ \tan(x)-x=\sum_{k=0}^\infty2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1})\tag{5} \] By \((3)\) each term of \((4)\) is between \(-\frac12-\epsilon\) and \(-\frac12+\epsilon\) of the corresponding term of \((5)\).

Следовательно, \[ \left|\,\frac{\sin(x)-x}{\tan(x)-x}+\frac12\,\right|\le\epsilon\tag{6} \] Thus, \[ \lim_{x\to0}\,\frac{\sin(x)-x}{\tan(x)-x}=-\frac12\tag{7} \] Furthermore, \[ \begin{align} \frac{\tan(x)-\sin(x)}{x^3} &=\tan(x)(1-\cos(x))\frac1{x^3}\\ &=\frac{\sin(x)}{\cos(x)}\frac{\sin^2(x)}{1+\cos(x)}\frac1{x^3}\\ &=\frac1{\cos(x)(1+\cos(x))}\left(\frac{\sin(x)}{x}\right)^3\tag{8} \end{align} \] Therefore, \[ \lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}=\frac12\tag{9} \] Combining \((7)\) and \((9)\) yield \[ \lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag{10} \] Additionally, \[ \frac{\sin(A)-\sin(B)}{\sin(A-B)} =\frac{\cos\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)} =1-\frac{2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}\tag{11} \]

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Finishing Up:

\[ \begin{align} &x\sin(\sin(x))-\sin^2(x)\\ &=[\color{#C00000}{(x-\sin(x))+\sin(x)}][\color{#00A000}{(\sin(\sin(x))-\sin(x))+\sin(x)}]-\sin^2(x)\\ &=\color{#C00000}{(x-\sin(x))}\color{#00A000}{(\sin(\sin(x))-\sin(x))}\\ &+\color{#C00000}{(x-\sin(x))}\color{#00A000}{\sin(x)}\\ &+\color{#C00000}{\sin(x)}\color{#00A000}{(\sin(\sin(x))-\sin(x))}\\ &=(x-\sin(x))(\sin(\sin(x))-\sin(x))+\sin(x)(x-2\sin(x)+\sin(\sin(x)))\tag{12} \end{align} \]

Using \((10)\), we get that

\[ \begin{align} &\lim_{x\to0}\frac{(x-\sin(x))(\sin(\sin(x))-\sin(x))}{x^6}\\ &=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{\sin(\sin(x))-\sin(x)}{\sin^3(x)}\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^3\\ &=\frac16\cdot\frac{-1}6\cdot1\\ &=-\frac1{36}\tag{13} \end{align} \] and with \((10)\) and \((11)\), we have \[ \begin{align} &\lim_{x\to0}\frac{\sin(x)(x-2\sin(x)+\sin(\sin(x)))}{x^6}\\ &=\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\frac{x-2\sin(x)+\sin(\sin(x))}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-(\sin(x)-\sin(\sin(x))}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))\left(1-\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}\right)}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))+\sin(x-\sin(x))\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}}{x^5}\\ &=\lim_{x\to0}\frac{\sin(x-\sin(x))}{x^3}\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{x^2}\\[6pt] &=\frac16\cdot\frac12\\[6pt] &=\frac1{12}\tag{14} \end{align} \]

Складывая \((13)\) и \((14)\)

получаем

\[\color{#C00000}{\lim_{x\to0}\frac{x\sin(\sin(x))-\sin^2(x)}{x^6}=\frac1{18}}\tag{15} \]

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Объяснение формулы  \((6)\)

Объяснение ниже
\(x\gt0\) and \(x\lt0\).

Только реверсируем (reverse the red inequalities).

Рассчитаем
\(x\color{#C00000}{\gt}0\) and \(|x|\lt\pi/2\). Then \(\tan(x)-2\tan(x/2)\color{#C00000}{\gt}0\).

\((3)\) эквивалентно
\[ \begin{align} &(-1/2-\epsilon)(\tan(x)-2\tan(x/2))\\[4pt] \color{#C00000}{\le}&\sin(x)-2\sin(x/2)\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)(\tan(x)-2\tan(x/2))\tag{16} \end{align} \] for all \(|x|\lt\delta\). Thus, for \(k\ge0\), \[ \begin{align} &(-1/2-\epsilon)(2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1}))\\[4pt] \color{#C00000}{\le}&2^k\sin(x/2^k)-2^{k+1}\sin(x/2^{k+1})\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)(2^k\tan(x/2^k)-2^{k+1}\tan(x/2^{k+1}))\tag{17} \end{align} \] Summing \((17)\) from \(k=0\) to \(\infty\) yields \[ \begin{align} &(-1/2-\epsilon)\left(\tan(x)-\lim_{k\to\infty}2^k\tan(x/2^k)\right)\\[4pt] \color{#C00000}{\le}&\sin(x)-\lim_{k\to\infty}2^k\sin(x/2^k)\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)\left(\tan(x)-\lim_{k\to\infty}2^k\tan(x/2^k)\right)\tag{18} \end{align} \] Since \(\lim\limits_{k\to\infty}2^k\tan(x/2^k)=\lim\limits_{k\to\infty}2^k\sin(x/2^k)=x\), \((18)\) says \[ \begin{align} &(-1/2-\epsilon)(\tan(x)-x)\\[4pt] \color{#C00000}{\le}&\sin(x)-x\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)(\tan(x)-x))\tag{19} \end{align} \] which, since \(\epsilon\)  эквивалентно \((6)\).