Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 40

Determine the limit $\lim\limits_{x \rightarrow -6} \displaystyle \frac{2x+1}{|x+6|}$, if it exists. If the limit does not exist, explain why.

The function contains an absolute value, therefore, we evaluate its left and right hand limit



$
\begin{equation}
\begin{aligned}
\text{For the right hand limit}\\
\lim\limits_{x \to -6^+} \frac{2x+12}{|x+6|} & = \lim\limits_{x \to -6^+} \frac{2x+12}{x+6}\\
\phantom{x} & = \lim\limits_{x \to -6^+} \frac{2\cancel{(x+6)}}{\cancel{(x+6)}}\\
\phantom{x} & = \lim\limits_{x \to -6^+} 2\\
\phantom{x} & = 2\\
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{For the left hand limit}\\
\lim\limits_{x \to -6^-} \frac{2x+2}{|x-6|} & = \lim\limits_{x \to -6^-} \frac{2x+12}{-(x+6)}\\
\phantom{x} & = \lim\limits_{x \to -6^-} \frac{2(x+6)}{-(x+6)}\\
\phantom{x} & = \lim\limits_{x \to -6^-} -2\\
\phantom{x} & = -2
\end{aligned}
\end{equation}
$


The left and right hand limits are different. Therefore, the limit does not exist.