Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 16

Determine the $\displaystyle \lim \limits_{x \to -1} \frac{x^2 - 4x}{x^2 - 3x - 4}$, if it exists.


$\begin{equation} \begin{aligned} & \lim \limits_{x \to -1} \frac{x^2 - 4}{x^2 - 3x - 4} = \lim \limits_{x \to -1} \frac{x \cancel{(x - 4)}}{\cancel{(x - 4)}(x +1)} && \text{ Get the factor and cancel out like terms. }\\ \\ & \lim \limits_{x \to -1} \frac{x}{x + 1} = \frac{-1}{-1 + 1} = \frac{-1}{0} && \text{ Result will be undefine }\\ \\ & \fbox{$\lim \limits_{x \to -1} \displaystyle \frac{x^2 - 4}{x^2 - 3x -4} \text{ Limit does not exist }$ } \end{aligned} \end{equation}$