Solution
= Solution
Factor out the dominant cubic term and use a <geometric series> in $1/(3x^2)$:
$$
\frac1{x+3x^3}=\frac1{3x^3}\left(1+\frac1{3x^2}\right)^{-1}=\frac1{3x^3}-\frac1{9x^5}+O(x^{-7}).
$$
Thus the requested leading term and absolute error order are
$$
\boxed{g(x)=\frac1{3x^3}+O(x^{-5})\qquad(x\to+\infty).}
$$
The relative error of retaining only the leading term is $O(x^{-2})$; it differs from the absolute error order $O(x^{-5})$.