Why does $\frac{\sin x}{x(x^2-4)}$ not have a vertical asymptote at $ x = 0 $ I suspect it has something to do with $$\lim_{x \to 0} \frac{\sin x}{x} = 1 $$ but I'm not too sure as I'm still using the precalc-learned rule where you find vertical asymptotes by finding $x$ values that makes the denominator of a function equal to $0$.
 A: Your suspicion is correct.
$$\frac{\sin x}{x(x^2-4)}=\frac{\sin x}{x}\cdot \frac{1}{x^2-4}.$$
$$\lim_{x\to0}\frac{\sin x}{x}=1,$$
as you mentioned, and
$$\lim_{x\to0}\frac{1}{x^2-4}=-\frac14,$$
by substitution.
The limit of a product of two functions is the product of the limits of the two functions (provided the limits exist) (limit product rule).
So
$$\lim_{x\to0}\frac{\sin x}{x(x^2-4)}=\lim_{x\to0}\frac{\sin x}{x}\times \lim_{x\to0}\frac{1}{x^2-4}=-\frac14.$$
The precalc rule that the vertical asymptotes are at the zeroes of the denominator work only if the zeroes of the denominator aren't also zeroes of the numerator. In this case, $x=0$ is both a root of the numerator and the donominator.
Consider the function $f(x)=\frac{(x-5)^2}{x-5}$ for example. $x=5$ is a zero of the denominator, but it is easy to see that there is no vertical asymptote; it simplifies to $f(x)=x-5$ for $x\neq5$. There is only a "hole" (a removable discontinuity) at $x=5$, just like there is at $x=0$ in the function in your question.
