Answer
$\frac{1}{120}$
Work Step by Step
$sinx=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....$
Plug into the limit to get
$\lim\limits_{x \to 0}\frac{sinx-x+1/6x^{3}}{x^{5}}=\lim\limits_{x \to 0}\frac{(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....)-x+1/6x^{3}}{x^{5}}$
$=\frac{1}{120}$
