Exactly when does a limit of a rational function exist? Let $f(x) = p(x)/q(x)$ be a rational function. Let q have a root at some $x = a$. It is clear that if $q(x)$ divides $p(x)$ then the limit of $f$ as $x$ goes to a exists. But is the converse true as well? 
 A: The fact that $q(x)$ divides $p(x)$ cannot be derived if
$$
\lim_{x\to a}\frac{p(x)}{q(x)}
$$
exists, be $a$ a root of $q(x)$ or not.
Obviously, if $a$ is not a root of $q(x)$ the limit exists, by continuity.
If $a$ is a root of $q(x)$, then you can write $q(x)=(x-a)^mq_1(x)$, where $a$ is not a root of $q_1(x)$: just divide by $x-a$ as many times as you can.
In this case $m>0$ by assumption. You can also write $p(x)=(x-a)^np_1(x)$, with $a$ not a root of $p_1(x)$; here it is possible that $n=0$ (when $a$ is not a root of $p(x)$ to begin with. Then we have
$$
\frac{p(x)}{q(x)}=\frac{(x-a)^n}{(x-a)^m}\frac{p_1(x)}{q_1(x)}
$$
and, since $q_1(a)\ne0$, the limit
$$
\lim_{x\to a}\frac{p_1(x)}{q_1(x)}
$$
exists finite. So we just need to look at
$$
\lim_{x\to a}\frac{(x-a)^n}{(x-a)^m} \tag{*}
$$
This exists finite if and only if $n\ge m$, because otherwise it can be rewritten as
$$
\lim_{x\to a}\frac{1}{(x-a)^{m-n}}
$$
which does not exist if $m-n$ is odd (it is $\infty$ from the right and $-\infty$ from the left) and is $\infty$ if $m-n$ is even.
If $n=m$, then the limit ($\text{*}$) is $1$; if $n>m$, then the limit ($\text{*}$) is $0$.
Just play with some rational function to understand what happens:
$$
\lim_{x\to 1}\frac{x^2+x-2}{x^2-1}=2
$$
but obviously $(x-1)(x+1)$ does not divide $(x-1)(x+2)$.
If you instead mean that the limit exists finite for all roots of $q(x)$, then the statement is false nonetheless. For instance
$$
\lim_{x\to1}\frac{x^2-1}{x^3-x^2+x-1}
$$
exists finite, and $1$ is the only (real) root of $x^3-x^2+x-1$.
A: The converse is not necessarily true. 
For example $$ \lim _{x\to 2} \frac {x^2-4}{x^2-5x+6} = -4 $$ 
But $(x^2-5x+6)$ does not divide $(x^2-4).$    
A: If $q(x)$ divides $p(x)$, the quotient will be a polynomial, so the converse is also true.
In general instead, if $x=a$ is a zero for $q(x)$, i.e a pole for the rational function, and $f(a)$ (in the limit) is finite, then it means that $x=a$ is also a zero for $p(x)$ with multiplicity equal or greater that that of $p(x)$ and vice-versa, when you also consider multiplicity to be null.
