Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 23

Determine the $\displaystyle \lim \limits_{x \to 7} \frac{\sqrt{x + 2} - 3}{x - 7}$, if it exists.


$
\begin{equation}
\begin{aligned}
\lim \limits_{x \to 7} \frac{\sqrt{x + 2} - 3}{x - 7} \cdot \frac{\sqrt{x + 2} + 3}{\sqrt{x + 2} + 3} &=
\lim \limits_{x \to 7} \frac{x + 2 - 9}{(x - 7)(\sqrt{x + 2} + 3)} = \lim \limits_{x \to 7} \frac{x - 7}{(x - 7)(\sqrt{x + 2} + 3)}
&& \text{ Multiply numerator and denominator by $(\sqrt{x + 2} + 3)$ then simplify. }\\
\\
& = \lim \limits_{x \to 7} \frac{\cancel{x - 7 }}{\cancel{x - 7 }(\sqrt{x + 2} + 3)} = \lim \limits_{x \to 7} \frac{1}{\sqrt{x +2 } + 3}
&& \text{ Cancel out like terms}\\
\\
&= \frac{1}{\sqrt{7 + 2 } + 3} = \frac{1}{\sqrt{9}+3} = \frac{1}{3+3}
&& \text{ Substitute value of $x$ and simplify}\\
\\
& = \frac{1}{6}

\end{aligned}
\end{equation}
$