Past exam of the mathematical methods II course of the University of Oxford / 2013 / 2 / a / Solution

Put $h=x-\pi /2$. Periodicity gives $\tan (2x)=\tan (\pi +2h)=\tan (2h)$. The local Taylor expansion $\tan (2h)=2h+O(h^3)$ therefore yieldsAlthough $\tan x$ is singular at $\pi /2$, the numerator here is $\tan (2x)$, which is smooth and zero there.