Remark 2

A function defined on a set in $\mathfrak{F}$ need not have a determinate order (finite or not) relative to g, for the functions having a determinate order relative to g are comparable to all the powers of g with at most one exception. Now f need not have this property, as one sees from the example $g(x) = x,\ f(x) = 1 + x^2 \sin^2 x$ (as x tends to $+\infty$). In this example f is comparable to $g^\alpha$ for $\alpha < 0$ and $\alpha > 2$; if one takes $f(x) = e^x \sin^2 x + e^{-x} \cos^2 x$ then f is not comparable to any power (positive or negative) of g.