Remark 2

Though $f$ and $g$ are comparable of order $k$ a function $f_1$ equivalent to $f$ need not be comparable of order $k$ to $g$; however it will be so if one assumes that $f_1$ is comparable of order $k$ to $f$ and that none of the derivatives $f^{(p)}$ ($0 \leqslant p \leqslant k - 1$) is equivalent to a nonzero constant.