If $f(x)$ and $g(x)$ are continuous at $x=a$, and
if $c$ is a constant, then

$f(x)+g(x)$ is continuous at $x=a$,

$f(x)-g(x)$ is continuous at $x=a$,

$cf(x)$ is continuous at $x=a$,

$f(x)g(x)$ is continuous at $x=a$, and

$\frac{f(x)}{g(x)}$ is continuous at $x=a$ as long as $g(a) \ne 0$.

These rules follow directly from the limit laws. For instance,
to see that $f(x)+g(x)$ is continuous at $x=a$, we need to show that
$\displaystyle{\lim_{x \to a} (f(x)+g(x))}=f(a)+g(a)$.
But
\begin{eqnarray*}
& \displaystyle{\lim_{x\to a} f(x)} = f(a) & \qquad \hbox{(since $f(x)$
is continuous),}\cr
& \displaystyle{\lim_{x\to a} g(x)} = g(a) & \qquad \hbox{(since $g(x)$ is continuous),}
\cr
& \displaystyle{\lim_{x \to a} (f(x) + g(x)) =
\lim_{x \to a} f(x) + \lim_{x \to a} g(x)} & \qquad \hbox{(by our limit laws),
so}\cr
& \displaystyle{\lim_{x \to a} (f(x) + g(x)) = f(a) + g(a)}, &
\end{eqnarray*}
as required. Deriving the other properties is similar.

Our next property involves composite functions:

Theorem: If $f(x)$ is continuous at $x=b$, and if
$\displaystyle{\lim_{x \to a}} g(x) = b$, then
$\displaystyle{\lim_{x \to a} f(g(x)) = f(b)}$.

To see this, suppose that $x$ is close to (but not equal) to $a$.
Then $g(x)$ is close to $b$, since $\displaystyle{\lim_{x \to a}g(x)=b}$.
Let $y=g(x)$. Since $f$ is continuous at $b$,
whenever $y$ is close to $b$, $f(y)$
is close to $f(b)$. But that makes $f(g(x))$ close to $f(b)$.

The following video goes over these properties and how to use them.