In the chapter singularity I found that if $f$ is analytic on $B'(a;r)$ and if either $(i)$ $\lim_{z \rightarrow a} |z-a|^{\alpha}|f(z)| = 0$ or $(ii)$ $\lim_{z \rightarrow a} |z-a|^{\alpha} |f(z)|=\infty$ holds for some $\alpha \in \mathbb R$ then there is an integral division of the real line i.e. there exists an integer $h$ such that $(i)$ holds for all $\alpha >h$ and $(ii)$ holds for all $\alpha <h$ so that $(i) \iff (ii)$. If $f$ has a pole at $a$ of order $k$ then the integral division is precisely $k$. If $f$ has a removable singularity at $a$ and $\lim_{z \rightarrow a} f(z)$ is a non-zero constant then clearly the integral division is $0$. Now if $\lim_{z \rightarrow a} f(z) = 0$ then what would be the integral division? Is it still $0$ or some other? According to me the integral division should exist because for removable singularity $(i)$ always holds for $\alpha \geq 1$ and hence by the above discussion it is clear that $(ii)$ must hold which justifies the existence of such integral division. Is it not so?
Please help me in understanding this concept,
Thank you in advance.