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?

Note,
$$\sin x\sin^{-1}x=(x-\frac{x^3}6 +\frac{x^5}{120}+O(x^7)) (x+\frac{x^3}6 +\frac{3x^5}{40}+O(x^7))=x^2+\frac{x^6}{18}+O(x^8)$$
Thus,
$$\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}
=\lim\limits_{x\to 0}\dfrac1{x^6} \left(\frac{x^6}{18}+O(x^8)\right)=\dfrac{1}{18}$$
