Solution
= Solution
The <Taylor series> of the <sine> and <arctangent> give $\sin x-\arctan x=x^3/6+O(x^5)$, while $x^2\ln(1+x)=x^3(1-x/2+O(x^2))$. Dividing the two series gives
$$
\boxed{\frac{\sin x-\arctan x}{x^2\ln(1+x)}=\frac16+O(x).}
$$
The first nonzero term after cancellation is the constant $1/6$.