Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 1 / b

The lower-order terms cancel, so the numerator must be expanded through sixth order. The Taylor series give $\sin (\sin x)=x-x^3/3+x^5/10+O(x^7)$ and ${\sin }^{2}x=x^2-x^4/3+2x^6/45+O(x^8)$. ThereforeThe first nonzero term is $x^2/18$, with the quotient continuously extended to zero if needed.