Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 12

Determine $\displaystyle \lim \limits_{x \to 3} \frac{\sqrt{x + 6} -x}{x^3 - 3x^2} $



$
\begin{equation}
\begin{aligned}

\lim \limits_{x \to 3} \frac{\sqrt{x + 6} -x}{x^3 - 3x^2} \cdot \frac{\sqrt{x + 6} + x}{\sqrt{x + 6} + x} &= \lim \limits_{x \to 3} \frac{x + 6 -x^2}{(x^3 - 3x^2)(\sqrt{x + 6 } + x)}
&& \text{Multiply both numerator and denominator by $(\sqrt{x + 6} + x)$}\\
\\
& = \lim \limits_{x \to 3} \frac{-\cancel{(x - 3)} (2 + x)}{x^2\cancel{(x - 3)}(\sqrt{x + 6} + x)}
&& \text{Factor numerator and denominator and cancel out like terms}\\
\\
&= \frac{-(2 + 3)}{(3)^2(\sqrt{3 + 6} + 3)} = \frac{-5}{(9)(6)}
&& \text{Substitute value of $x$ and simplify}\\
\\
& \fbox{$= \displaystyle \frac{-5}{54}$}

\end{aligned}
\end{equation}
$