A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$

Preliminary Results:

We will use $$ \begin{align} \frac{\color{#C00000}{\sin(2x)-2\sin(x)}}{\color{#00A000}{\tan(2x)-2\tan(x)}} &=\underbrace{\color{#C00000}{2\sin(x)(\cos(x)-1)}\vphantom{\frac{\tan^2(x)}{\tan^2(x)}}}\underbrace{\frac{\color{#00A000}{1-\tan^2(x)}}{\color{#00A000}{2\tan^3(x)}}}\\ &=\hphantom{\sin}\frac{-2\sin^3(x)}{\cos(x)+1}\hphantom{\sin}\frac{\cos(x)\cos(2x)}{2\sin^3(x)}\\ &=-\frac{\cos(x)\cos(2x)}{\cos(x)+1}\tag{1} \end{align} $$ Therefore, $$ \lim_{x\to0}\frac{\sin(x)-2\sin(x/2)}{\tan(x)-2\tan(x/2)}=-\frac12\tag{2} $$ Thus, given an $\epsilon\gt0$, we can find a $\delta\gt0$ 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 $(5)$. Therefore, $$ \left|\,\frac{\sin(x)-x}{\tan(x)-x}+\frac12\,\right|\le\epsilon\tag{6} $$ 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, $$ \frac{\sin(A)-\sin(B)}{\sin(A-B)} =\frac{\cos\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)} =1-\frac{2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)}\tag{11} $$

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 $$ \begin{align} &\lim_{x\to0}\frac{\sin(x)(x-2\sin(x)+\sin(\sin(x)))}{x^6}\\ &=\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\frac{x-2\sin(x)+\sin(\sin(x))}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-(\sin(x)-\sin(\sin(x))}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))\left(1-\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}\right)}{x^5}\\ &=\lim_{x\to0}\frac{(x-\sin(x))-\sin(x-\sin(x))+\sin(x-\sin(x))\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{\cos\left(\frac{x-\sin(x)}{2}\right)}}{x^5}\\ &=\lim_{x\to0}\frac{\sin(x-\sin(x))}{x^3}\frac{2\sin\left(\frac{x}{2}\right)\sin\left(\frac{\sin(x)}{2}\right)}{x^2}\\[6pt] &=\frac16\cdot\frac12\\[6pt] &=\frac1{12}\tag{14} \end{align} $$ Adding $(13)$ and $(14)$ gives $$ \color{#C00000}{\lim_{x\to0}\frac{x\sin(\sin(x))-\sin^2(x)}{x^6}=\frac1{18}}\tag{15} $$

Added Explanation for the Derivation of $(6)$

The explanation below works for $x\gt0$ and $x\lt0$. Just reverse the red inequalities.

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

$(3)$ is equivalent to $$ \begin{align} &(-1/2-\epsilon)(\tan(x)-2\tan(x/2))\\[4pt] \color{#C00000}{\le}&\sin(x)-2\sin(x/2)\\[4pt] \color{#C00000}{\le}&(-1/2+\epsilon)(\tan(x)-2\tan(x/2))\tag{16} \end{align} $$ 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 $\epsilon$ is arbitrary is equivalent to $(6)$.

$\eqalign{ L \;&=\; \lim\limits_{x \to 0} \dfrac {x-\sin x} {x^3} \\ \;&=\; \lim\limits_{t \to 0} \dfrac {3t-\sin 3t} {27t^3}\qquad x=3t \\ \;&=\; \lim\limits_{t \to 0} \dfrac{3t-3\sin t+4\sin^3t}{27t^3}\qquad \sin3t=3\sin t-4\sin^3t \\ \;&=\; \dfrac19\lim\limits_{t \to 0} \dfrac{t-\sin t}{t^3}+\dfrac4{27}\lim\limits_{t\to0}\dfrac{\sin^3t}{t^3} \\ \;&=\; \dfrac L9+\dfrac4{27} \\ \;&\Rightarrow\;\;\,\text{we obtain }\, L=\dfrac16 }$

I could prove it without using L'Hospital's rule, though I needed the following formula for $\sin x$ $$\sin{x}=x\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)=x\left(1-\frac{x^2}{6}+\frac{x^4}{120}+O(x^6)\right)$$ and the observation $$\sin ^2x=x^2\left(1-\frac{x^2}{3}+O(x^4)\right)$$ The constants $1/6$ and $1/120$ are due to $\zeta(2)/\pi^2$ and $\frac{1}{2}(\zeta^2(2)-\zeta(4))$ respectively. I also have used the simple formula $$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$

Now I start the proof \begin{equation} \begin{split} \ &\lim_{x\rightarrow 0}\frac{x\sin(\sin x)-\sin^2x}{x^6}\\ \ = & \lim_{x\rightarrow 0}\frac{x\sin x \left(\prod_{k=1}^\infty\left(1-\frac{\sin^2x}{k^2\pi^2}\right)\right)-\sin^2x}{x^6}\\ \ =&\lim_{x\rightarrow 0}\frac{\sin x}{x} \lim_{x\rightarrow 0}\frac{x \left(\prod_{k=1}^\infty\left(1-\frac{\sin^2x}{k^2\pi^2}\right)\right)-x\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)}{x^5}\\ \ =&\lim_{x\rightarrow 0}\frac{(1-\frac{\sin^2x}{6}+\frac{\sin^4 x}{120}+O(\sin^6x))-(1-\frac{x^2}{6}+\frac{x^4}{120}+O(x^6))}{x^4}\\ \ =&\lim_{x\rightarrow 0}\frac{1-\frac{\sin^2x}{x^2}}{6x^2}+\lim_{x\rightarrow 0}\frac{\sin^4x-x^4}{120x^4}\\ \ =&\lim_{x\rightarrow 0}\frac{1-(1-\frac{x^2}{3}+O(x^4))}{6x^2}+\frac{1}{120}(1-1)\\ \ =&\frac{1}{18}\hspace{0.6cm} \Box \end{split} \end{equation}