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.


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