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}
$