Now, $$ \lim_{x\to 0} \dfrac{sinx \cdot sin^{-1}x - x^2}{x^6} \;=\; \lim_{x\to 0} \dfrac{ (x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \ldots) (x + \dfrac{x^3}{6} + \dfrac{3x^5}{40} - \ldots) -x^2}{x^6} \\ = \lim_{x\to 0} \dfrac{ x^2 + \dfrac{x^4}{6} + \dfrac{3x^6}{40} - \dfrac{x^4}{6} - \dfrac{x^6}{36} + \dfrac{x^6}{120} + (Higher \; powers \; of \; x \; more \; than \; 6) - x^2 }{x^6} $$

$$ \therefore \lim_{x\to 0} \dfrac{sinx \cdot sin^{-1}x - x^2}{x^6} \;=\; \lim_{x\to 0} \dfrac{ \dfrac{27x^6 - 10x^6 + 3x^6}{360} }{x^6} \;=\; \dfrac{1}{18} $$