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}$