$$\lim_{x \to 0} \frac{x\sin(\sin x) - \sin^2 x}{x^6}$$

**My Attempt: **

I started with L'Hopital's rule. But it quickly became messy. So, I did not continue.

I tried to write the Taylor series of $\sin x$ to see if that $x^6$ gets canceled anywhere. But got stuck because of that $\sin(\sin x)$ term.

I have no Idea how do I approach this problem further.

Any help would be appreciated.


Note that\begin{align}\lim_{x\to0}\frac{x\sin(\sin x)-\sin^2x}{x^6}&=\lim_{x\to0}\frac{\arcsin(\sin x)\sin(\sin x)-\sin^2x}{\arcsin^6(\sin x)}\\&=\lim_{y\to0}\frac{\arcsin(y)\sin(y)-y^2}{\arcsin^6y}\\&=\frac1{18},\end{align}because\begin{align}\arcsin(y)\sin(y)&=\left(y+\frac{y^3}6+\frac{3y^5}{40}+\cdots\right)\left(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots\right)\\&=y^2+\frac{y^6}{18}+\cdots,\end{align}whereas $\arcsin^6y=y^6+\cdots$


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