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

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.

Combine the terms before taking their limit:Using the exponential function Maclaurin series, the numerator is $x^2/2+O(x^3)$ and the denominator is $x^2+O(x^3)$. ConsequentlyThe cancellation removes the two individual singularities. Subtracting their separate divergent limits would not determine the finite answer.