$\lim_{z\to z_0} \frac{p(z)}{q(z)}$ where $p$ and $q$ are complex polynomials. Let $p:\mathbb{C}\to \mathbb{C}$ defined by $p(z) = a_0+a_1z+\cdots + a_nz^n$ and $q:\mathbb{C}\to \mathbb{C}$ defined by $q(z) = b_0+b_1z+\cdots + b_nz^n$, take:
$$\lim_{z\to z_0} \frac{p(z)}{q(z)}$$
For which points $z_0$ we can calculate this limit?
I thought about points that are not in the set of roots of $q(z)$. Is there anymore things I must consider?
 A: Hint 
Suppose $p(z)=(z-z_0)^\alpha \ a(z)$ and $q(z)=(z-z_0)^\beta \ b(z)$ where $a(z_0),b(z_0)$ are nonzero. Then the limit exists iff $\alpha \geq \beta$
Explanation
$$
\lim\limits_{z\rightarrow z_0} \frac{p(z)}{q(z)} 
\ = \ \lim\limits_{z\rightarrow z_0}(z-z_0)^{\alpha-\beta}\lim\limits_{z\rightarrow z_0}\frac{a(z)}{b(z)} 
\ = \  \lim\limits_{z\rightarrow z_0}(z-z_0)^{\alpha-\beta}\frac{a(z_0)}{b(z_0)}
$$
The last limit exists iff $\alpha\geq \beta$.
Note
Here $\alpha$ and $\beta$ are nonnegative integers. If $p(z_0)\neq 0$ then we take $\alpha=0$. If $p(z_0) = 0$ then we take the greatest integer $\alpha$ such that $(z-z_0)^\alpha$ divides $p(z)$ or equivalently $\frac{p(z)}{(z-z_0)^\alpha}$ is a polynomial.
This is the usual factorization to do for polynomials. 
A: You may use Euclidean division to find the greatest common divisor $d(z)$ of $p(z)$ and $q(z)$. This gives you $p(z)=r(z)d(z)$ and $q(z)=s(z)d(z)$ where $r$ and $s$ have no roots in commun.
Then
$$ \lim_{z\rightarrow z_0}\frac{p(z)}{q(z)} =\lim_{z\rightarrow z_0} \frac{r(z)}{s(z)}$$
is infinity precisely when $s(z_0)=0$.
