Calculus and Its Applications, Chapter 1, 1.2, Section 1.2, Problem 26

Determine the $\displaystyle \lim_{x \to -4^-} \sqrt{x^2 - 16}$ by using the Limit Principles.
If the limit does not exist, state the fact.


$
\begin{equation}
\begin{aligned}
\lim_{x \to -4^-} \sqrt{x^2 - 16} &= \sqrt{\lim_{x \to -4^-} (x^2 - 16)}
&& \text{The limit of a root is the root of the limit}\\
\\
&= \sqrt{\lim_{x \to -4^-} x^2 - \lim_{x \to -4^-} 16}
&& \text{The limit of a difference is the difference of the limits}\\
\\
&= \sqrt{\left(\lim_{x \to -4^-} x \right)^2 - \lim_{x \to -4^-} 16}
&& \text{The limit of a power is the power of the limit}\\
\\
&= \sqrt{\left(\lim_{x \to -4^-} x \right)^2 - 16}
&& \text{The limit of a constant is the constant}\\
\\
&= \sqrt{(-4)^2 -16}
&& \text{Substitute }-4\\
\\
&= \sqrt{16 - 16}\\
\\
&= 0
\end{aligned}
\end{equation}
$