A rational function is a function $f$ of the form $p/q$ where $p$ and $q$ are polynomial functions. The domain of $f$ is $\{x\in \mathbb R: q(x) \neq 0\}$. Prove every rational function is continuous.
I have previously proved that every polynomial function $p(x)=a_0+a_1x+...+...a_nx^n$ is continuous on $\mathbb R$. Could I use a theorem that states $f/g$ is continuous at $x_0$ if $g(x_0) \neq 0$? Then $p(x)$ (which is already proved to be continuous) over $q(x)$ would be continuous?
A function $f$ of $x$ is continuous at $x=a$ if and only if $$\lim_{x\to a}f(x)=f(a)$$
Obviously, as you’ve stated, not every rational function is continuous on the entire real line, so you just need to prove there is some interval $I\subset\Bbb{R}$ on which $f$ is continuous—that is, there is an $a\in I$ that satisfies the above criterion.
So argue that there is at least one $a\in\Bbb{R}\setminus\{x:q(x)=0\}$ such that
$$\begin{align} \lim_{x\to a}\frac{p(x)}{q(x)}&=\frac{p(a)}{q(a)}\\ \frac{\lim_{x\to a}p(x)}{\lim_{x\to a}q(x)}&=\frac{p(a)}{q(a)}\\ \frac{\lim_{x\to a}p(x)}{\lim_{x\to a}q(x)}&\in\Bbb{R}\\ \end{align}$$
Since there is only a finite number of $x:q(x)=0$ and since $p(x)$ is defined across the whole real line, then there are an infinite number of valid $a$.
