Past exam of the mathematical methods II course of the University of Oxford / 2015 / 2 / a

The Maclaurin series $\sin x=x-x^3/6+O(x^5)$ cancels the numerator's linear term, leavingThus $\boxed{\underset{x\to 0}{\lim }(\sin x-x)/x^3=-1/6}$. The same expansion applies for positive and negative $x$, so this is a two-sided limit.