<s>[INST] Show that $\lim\limits_{x\to 0} \frac{\sin x\sin^{-1}x-x^2}{x^6}=\frac1{18}$ Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$ 
My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\sin^{-1}x-x}{x^5}=\lim\limits_{x\to 0} \dfrac{\frac{1}{\sqrt{1-x^2}}-1}{5x^4}$ .
Is my approach correct?
 [/INST] When $x$ is small, $\sin x \approx x-x^3/6+x^5/120$ and $\sin^{-1} x\approx x+x^3/6+3x^5/40$
Then $$\lim_{x \rightarrow 0} \frac{(x-x^3/6+x^5/120)(x+x^3/6+3x^5/40)-x^2}{x^{6}}=
\lim_{x\rightarrow 0}\frac{x^6/18+(.)x^8}{x^6}=\frac{1}{18}.$$
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