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}
$