I'm aware there have been questions similar to this one; I've read them and my doubt persists, due to the fact that, as I said, those questions are similar, but not equal to the one that troubles me.
My doubt is very simple. Given a polynomial $Q(x)$ such that $Q(a)=0$, we need to find the solution for
$\lim_{x\to a}$ $\frac{b}{Q(x)}$
where $a, b \in \mathbb{Z}$.
As stated before, $Q(a) = 0$. On my collegue text-book, on certain pages we are told that, since when we evaluate our limit's argument on $a$ our denominator is $0$, our limit is undefined. On other parts it says this equals $\infty$, since our expression $\frac{b}{Q(x)}$ becomes increasingly smaller as $x$ approaches $a$. Which is the right answer, and where am I confussed? I'm extremely new to limits (last night was my first contact), so excuse me if this seems obvious for some of you folks.
Over $\mathbb{C},$ there is no confusion whatsoever. As written, as $a$ is a root of $Q(z)$, it follows that $a$ is a pole of $\frac{b}{Q(z)}$ hence the limit is by definition $\infty$, the complex infinity. Note that there is no concept of $+\infty$ or $-\infty$ over $\mathbb{C}$, as there is no order over the complex numbers, i.e. infinity looks the same from all directions.
Over $\mathbb{R}$, both answers in your textbook are possible. This is because over $\mathbb{R}$, a limit $\lim_{x\to a} f(x)$ exists iff the right and left limits $\lim_{x\to a+} f(x)$ and $\lim_{x\to a-} f(x)$ exist. Depending on the behavior of a given polynomial $p(x)$ in a small interval around $a$, it could be the case that both the left and right limits agree, in which case the limit is either $\infty$ or $-\infty$. It could also be the case that the left and right limits disagree, in which case we say the limit does not exist. I give examples of both scenarios below:
Suppose $p(x) = x, a=0, b=1$ , then we have $\lim_{x\to 0-} \frac{1}{x} = -\infty$ as $\frac{1}{x}$ takes strictly negative values for $x<0$. Similarily, $\lim_{x\to 0+} \frac{1}{x} = +\infty$ as $\frac{1}{x}$ takes strictly positive values for $x>0$. Thus, in this case the limit does not exist.
Suppose now $p(x) = x^2 ,a=0, b=1$, then $\lim_{x\to 0-} \frac{1}{x^2} = \lim_{x\to 0+} \frac{1}{x^2} = \infty$, as $\frac{1}{x^2}$ takes positive values for all $x\neq 0$. In this case the limit exists, and is equal to $\infty$.