Calculus and Its Applications, Chapter 1, 1.2, Section 1.2, Problem 22

Determine the $\displaystyle \lim_{x \to 5} \sqrt{x^2 - 16}$ by using the Limit Principles.
If the limit does not exist, state the fact.


$
\begin{equation}
\begin{aligned}
\lim_{x \to 5} \sqrt{x^2 - 16} &= \sqrt{\lim_{x \to 5} (x^2 - 16)}
&& \text{The limit of a root is the root of the limit}\\
\\
&= \sqrt{\lim_{x \to 5} x^2 - \lim_{x \to 5} 16}
&& \text{The limit of a difference is the difference of the limits}\\
\\
&= \sqrt{\left(\lim_{x \to 5} x \right)^2 - \lim_{x \to 5} 16}
&& \text{The limit of a power is the power of the limit}\\
\\
&= \sqrt{\left(\lim_{x \to 5} x \right)^2 - 16}
&& \text{The limit of a constant is the constant}\\
\\
&= \sqrt{5^2 - 16}
&& \text{Substitute }5\\
\\
&= \sqrt{25 - 16}\\
\\
&= \sqrt{9}\\
\\
&= 3
\end{aligned}
\end{equation}
$