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

Let $h=x-\pi /2$. The trigonometric identity $\sin (\pi /2+h)=\cos h$ changes the expression to $h^2/(\cos h-1)$. Since $\cos h-1=-h^2/2+O(h^4)$,The negative sign is also forced geometrically: near its maximum, $\sin x-1$ is negative while the numerator is nonnegative. Both numerator and denominator vanish quadratically, so one application of a first-derivative argument alone would not yet resolve the indeterminate form.