Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 30

Determine the $\displaystyle \lim \limits_{x \to -4} \frac{\sqrt{x^2 + 9} - 5}{x + 4}$, if it exists.


$
\begin{equation}
\begin{aligned}
& \lim \limits_{x \to -4} \frac{\sqrt{x^2 + 9} - 5}{x + 4}
\cdot \frac{\sqrt{x^2 + 9} + 5}{x + 4}
= \lim \limits_{x \to -4} \frac{x^2 + 9 -25}{(x + 4)(\sqrt{x^2 + 9} + 5)}
&& \text{ Multiply both numerator and denominator by $\sqrt{x^2 + 9} + 5$ then simplify.}\\
\\
& \lim \limits_{x \to -4} \frac{x^2 - 16}{(x + 4) (\sqrt{x^2 + 9} + 5)}
=\lim \limits_{x \to -4} \frac{\cancel{(x + 4)}(x - 4)}{\cancel{(x + 4)}(\sqrt{x^2 + 9} + 5)}
&& \text{ Get the factor and cancel out like terms.}
\\
& \lim \limits_{x \to -4} \frac{x - 4}{\sqrt{x^2 + 9} + 5}
= \frac{-4 - 4}{\sqrt{(-4)^2 + 9} + 5}
= \frac{-8}{\sqrt{16+9}+5}
= \frac{-8}{5+5}
= \frac{-8}{10}
= -\frac{4}{5}
&& \text{ Substitute value of $x$ and simplify}\\
\\
& \fbox{$ \lim \limits_{x \to -4} \displaystyle \frac{\sqrt{x^2 + 9} - 5}{x + 4} = -\frac{4}{5}$}


\end{aligned}
\end{equation}
$