Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 9

Determine the $\lim\limits_{x \rightarrow 4^-} \quad \sqrt{16-x^2}$ and justify each step by indicating the appropriate limit law(s).


$\begin{equation} \begin{aligned} \lim\limits_{x \rightarrow 4^-} \quad \sqrt{16-x^2} &= \sqrt{\lim\limits_{x \rightarrow 4^-} (16-x^2)} && \text{(Root Law)}\\ \lim\limits_{x \rightarrow 4^-} \quad \sqrt{16-x^2} &= \sqrt{\lim\limits_{x \rightarrow 4^-} 16 - \lim\limits_{x \rightarrow 4^-} x^2} && \text{(Difference Law)}\\ \lim\limits_{x \rightarrow 4^-} \quad \sqrt{16-x^2} &= \sqrt{16 - \lim\limits_{x \rightarrow 4^-}x^2 } && \text{(Constant Law)}\\ \lim\limits_{x \rightarrow 4^-} \quad \sqrt{16-x^2} &= \sqrt{16-(4)^2} && \text{(Power Special Limit Law)} \end{aligned} \end{equation}\\ \boxed{\lim\limits_{x \rightarrow 4^-} \quad \sqrt{16-x^2} = 0}$