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

The Maclaurin series give $1-\cos (x^2)=x^4/2+O(x^8)$ and $\tan x=x+O(x^3)$. Therefore $x^2{\tan }^{2}x=x^4+O(x^6)$, and cancellation of the common leading power givesThe argument applies from either side of zero: both leading terms involve the even power $x^4$. The expression need not be defined at zero itself for its limit to exist.

Combine the two singular terms before taking the limit:For $x$ near zero, within the real domain $x>-1$, the Maclaurin series is $\log (1+x)=x-x^2/2+O(x^3)$. HenceThe two individual terms have divergent limits; their leading singularities cancel. Evaluating them separately as an “infinity minus infinity” subtraction would not justify the finite result.

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.