\lim_{x\to 0}\frac{x-sinx}{x^{3}}=\frac{4}{3^{3}}+\frac{4}{3^{5}}+\frac{4}{3^{7}}+\frac{4}{3^{9}}+....=\left ( \frac{a_{1}}{1-q} \right )=\frac{\frac{4}{3^{3}}}{1-\frac{1}{9}}=\frac{\frac{4}{27}}{\frac{8}{9}}=\frac{1}{6}