Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 14

Show that the function $\displaystyle g(x) = 2\sqrt{3-x}$ is continuous on the interval $(-\infty,3]$ by using the definition
of continuity and the properties of limits.

By using the properties of limit, let's pick $a=2$ on the interval $(-\infty,3]$


$
\begin{equation}
\begin{aligned}
\lim\limits_{x \to 2} 2\sqrt{3-x} & = 2 \sqrt{ \lim\limits_{x \to 2} 3- \lim\limits_{x \to 2} x }
&& \text{(Applying Difference, Sum and Quotient Law.)}\\
& = 2\sqrt{3-2}
&& \text{(Substitute } a = 2)\\
& = 2
&& \text{(It shows that the function is continuous at 2 and is equal to 2)}
\end{aligned}
\end{equation}
$


By using the definition of continuity,
The given function is a rational function that is continuous at every number in its domain according to the theorem.
And the domain of the function is $(-\infty,3]$

Therefore,
The function is continuous on the interval $(-\infty,3]$