I'm studying the Levelt filtration and a certain valuation function comes up and I'm trying to understand when (and why) it is finite.
Let $S$ be a disk in the complex plane centred at $0$, $S' = S - 0$, and $W$ the universal cover of $S'$ (the Riemann surface of the logarithm). Given a meromorphic function $f$ defined on $W$ we say that it has finite order if there is some $\lambda \in \mathbb{R}$ such that $\frac{f(z)}{z^\lambda} \to 0$ as $z \to 0$ in every sector of $W$ of opening less than $2 \pi$. In such a case, define the valuation of $f$, $\phi(f)$, to be the largest integer $N$ for which the above limit holds for all $\lambda < N$.
For example, for $f = \sqrt{z}$ , $\phi(f) = 0$, and for $f = 0$, $\phi(f) = \infty$.
I would like to know in which cases is $\phi(f)$ finite (assuming that $f$ is of finite order). $\phi(f)$ can fail to be be finite if $f(z)$ approaches $0$ faster than any polynomial, such as is the case for $f(z) = \exp(-\frac{1}{z})$ in the right half plane. However, this function does not have finite order because of its behaviour in the left half plane. So I would like to know whether its possible for a nonzero meromorphic function to approach $0$ faster than any polynomial in any sector.