# Solve the following limit as $\lim_{x \to 0}$ [duplicate]

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

**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.

• Your power series approach is the way to go. Just put the first few terms of the series for $\sin x$ into the first few terms of the same series. You won't need too many. – Ethan Bolker Apr 22 at 15:00
• The limit should be $$\frac{1}{18}$$ – Dr. Sonnhard Graubner Apr 22 at 15:02
• – Martin R Apr 22 at 15:03
• socratic.org/questions/… – lab bhattacharjee Apr 22 at 15:04

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 $$\arcsin^6y=y^6+\cdots$$