The following theorems give us an easy way to determine if a
complicated function is continuous. We simply break the
function into simpler functions, and if our function is the
sum/difference/product/quotient/composition of continuous functions,
then it is continuous.

Theorem: If $f(x)$ and $g(x)$ are continuous at
$x=a$, and if $c$ is a constant, then $f(x)+g(x)$,
$f(x)-g(x)$, $cf(x)$, $f(x)g(x)$, and $\frac{f(x)}{g(x)}$
(if $g(a)\ne 0$) are continuous at $x=a$.

In short: the sum, difference, constant multiple,
product and quotient of continuous functions are
continuous.
(to understand why, see *
below)

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

In short: the composition of continuous functions is
continuous.
(to understand why, see **
below)

Theorem: polynomial,
rational, root, trigonometric, inverse trigonometric,
exponential, and logarithmic functions are continuous
at every number in their domain.

Why do we care? By
definition, if $f$ is continuous at $x=a$, then
$\displaystyle{\lim_{x\rightarrow a}f(x)=f(a)}$, which (if you think about it for a minute, and please DO)
means that to evaluate the limit of a
continuous function $f$ as $x\to a$, you need only plug in $a$
to $f$. It is easy to take the limit of a
continuous function!

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

* This
theorem is a direct result of 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{array}{lll}
\displaystyle\lim_{x\to a} f(x) &= f(a) &\hbox{(since
$f(x)$ is continuous),}\\
\displaystyle\lim_{x\to a} g(x) & = g(a) & \hbox{(since
$g(x)$ is continuous),} \\
\displaystyle\lim_{x \to a} (f(x) + g(x)) &\displaystyle=
\lim_{x \to a} f(x) + \lim_{x \to a} g(x) & \hbox{(by our
limit laws), so}\\
\displaystyle\lim_{x \to a} (f(x) + g(x)) &= f(a) + g(a),
\end{array}
as required. Deriving the other properties is similar.

** 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 $f(b)$.