Solution
= Solution
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 $\sin^2x=x^2-x^4/3+2x^6/45+O(x^8)$. Therefore
$$
\frac{x\sin(\sin x)-\sin^2x}{x^4}=\frac{x^2}{18}+O(x^4).
$$
The \b[first nonzero term is $x^2/18$], with the quotient continuously extended to zero if needed.