Evaluating $\lim_{x\to 0}\frac{\sin(x)\arcsin(x)-x^2}{x^6}$ Step by Step Using L' Hopital Rule The limit to be found is
$$ \lim_{x\to 0}\frac{\sin(x)\arcsin(x)-x^2}{x^6}$$
I've tried l'hopital rule but it gets really messy. I've also tried splitting it into 2 limits but that doesn't work. I can't think of any meaningful substitution either.
PS: I know the answer $1/18$ but I'm interested in the method. Thank you.
 A: If you know that the answer is $\frac 1  {18}$, there is probably a typo in the expression.
Any way consider $$\dfrac{\sin (x)\sin^{-1} (x^{2})}{x^{6}}=\frac 1 {x^3} \times \frac {\sin(x)} {x}\times \frac {\sin^{-1}(x^2)} {x^2}$$ and remember how behave the second and third term.
Edit
Now, the problem is $$\frac{\sin(x)\arcsin(x)-x^2}{x^6}$$ So, just as Workaholic commented, use the classical Taylor series $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+O\left(x^6\right)$$ Making the product $$\sin(x)\sin^{-1}(x)=x^2+\frac{x^6}{18}+O\left(x^7\right)$$
A: Use standard equivalents: $\sin x\sim_0 x,\quad\arcsin u\sim_0 u$, hence
$$\frac{\sin x\arcsin x^2}{x^6}\sim_0\frac{x\cdot x^2}{x^6}=\frac1{x^3}$$
Thus $\;\displaystyle\lim_{x\to 0^+}\frac{\sin x\arcsin x^2}{x^6}=+\infty$, $\quad\displaystyle\lim_{x\to 0^-}\frac{\sin x\arcsin x^2}{x^6}=-\infty$.
A: NOTE.-This answer was for the question before which is distinct of the given now
It is well known that in a circle of radius $r$ the arc $\widehat{AB}$ subtended by an angle $x$ is given by $\widehat {AB}=rx$. In the unitary circle it is also known that $\sin x\approx x$ for small values so in the figure $\overline{BB'}\approx \widehat{AB}$ for small values. From this it is deduced that, for all natural n (in particular for $n=2$) $$\lim\limits_{x\to 0} \frac{\arcsin x^n}{x^n}=1 $$

Writing your expression as the product $$\frac{\sin x}{x}\cdot \frac{\arcsin x^2}{x^2}\cdot \frac{1}{x^3} $$ you can see it is not possible your limit be  $\frac{1}{18}$ (It does not exist because "tends to" $\pm\infty$).I conclude that your $\frac {1}{18}$ as answer is a typo.
A: Some basic limits can be obtained by repeated use of
l'Hospital's Rule easily, 
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \ \ \ \
and\ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\arcsin x-x}{x^{3}}=\frac{1}{%
6} \\
\lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} &=&\frac{1}{120}%
,\ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\arcsin x-x-%
\frac{1}{6}x^{3}}{x^{5}}=\frac{3}{40}
\end{eqnarray*}
Now, re-write the original expression as follows: 
\begin{eqnarray*}
\frac{\sin x\arcsin x-x^{2}}{x^{6}} &=&\frac{(\left[ \sin x-x\right]
+x)([\arcsin x-x]+x)-x^{2}}{x^{6}} \\
&=&\frac{\left[ \sin x-x\right] [\arcsin x-x]+x\left[ \sin x-x\right]
+x[\arcsin x-x]+x^{2}-x^{2}}{x^{6}} \\
&=&\frac{\left[ \sin x-x\right] [\arcsin x-x]}{x^{6}}+\frac{x\left[ \sin x-x+%
\frac{1}{6}x^{3}\right] +x[\arcsin x-x-\frac{1}{6}x^{3}]}{x^{6}} \\
&=&\frac{\left[ \sin x-x\right] }{x^{3}}\frac{[\arcsin x-x]}{x^{3}}+\frac{%
\left[ \sin x-x+\frac{1}{6}x^{3}\right] }{x^{5}}+\frac{[\arcsin x-x-\frac{1}{%
6}x^{3}]}{x^{5}}
\end{eqnarray*}
It follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x\arcsin x-x^{2}}{x^{6}} &=&\lim_{x%
\rightarrow 0}\frac{\left[ \sin x-x\right] }{x^{3}}\lim_{x\rightarrow 0}%
\frac{[\arcsin x-x]}{x^{3}}+\lim_{x\rightarrow 0}\frac{\left[ \sin x-x+\frac{%
1}{6}x^{3}\right] }{x^{5}}+\lim_{x\rightarrow 0}\frac{[\arcsin x-x-\frac{1}{6%
}x^{3}]}{x^{5}} \\
&=&\left( -\frac{1}{6}\right) \left( \frac{1}{6}\right) +\left( \frac{1}{120}%
\right) +\left( \frac{3}{40}\right)  \\
&=&\frac{1}{18}.
\end{eqnarray*}
