<s>[INST] 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.
 [/INST] 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*}
</s>