Proposition 7

If two real functions $f,\ g$, are comparable of order 1, then they are comparable; further, the relation $f \ll g$ (resp. $f \sim cg$, $c$ constant) implies $f' \ll g'$ (resp. $f' \sim cg'$) unless $g$ is equivalent to a nonzero constant.

Now, since $f'$ and $g'$ are of constant sign on an interval $[x_0, +\infty[$, both $f$ and $g$ are monotone on this interval, so tend to a finite or infinite limit as $x$ tends to $+\infty$. It is clear that $f$ and $g$ are comparable as $x$ tends to $+\infty$ if one of these limits is finite and $\neq 0$, or if one is zero and the other infinite. If both $f$ and $g$ tend to 0 one can write $f(x) = -\int_x^{+\infty} f'(t) \, dt,\ g(x) = -\int_x^{+\infty} g'(t) \, dt$; since $f'$ and $g'$ are comparable the same is true for $f$ and $g$ and the comparison relation between $f$ and $g$ is the same as that between $f'$ and $g'$, by prop. 6 (V, p. 20). Similarly, if $f$ and $g$ both have an infinite limit one has $f(x) = f(x_0) + \int_{x_0}^x f'(t) \, dt,\ g(x) = g(x_0) + \int_{x_0}^x g'(t) \, dt$; again prop. 6 (V, p. 230) shows that $f$ and $g$ are comparable and that the comparison relation between $f$ and $g$ is the same as that between $f'$ and $g'$. To complete the proof it remains to consider the case where $g$ tends to $\pm \infty$ and $f$ to a constant: one cannot then have $f' \succcurlyeq g'$, since then one could deduce from prop. 1 (V, p. 218) that the integral $\int_{x_0}^{+\infty} g'(t) \, dt$ was convergent; since $f'$ and $g'$ have been assumed comparable, one must have $f' \ll g'$.