Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 39

Determine the limit $\lim\limits_{x \rightarrow 3}(2x+|x-3|)$, 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 3^+} (2x+|x-3|) & = \lim\limits_{x \to 3^+} (2x+x-3)\\
\phantom{x} & = \lim\limits_{x \to 3^+} (3x-3)\\
\phantom{x} & = 3(3)-3\\
\phantom{x} & = 6\\
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{For the left hand limit}\\
\lim\limits_{x \to 3^-} (2x+|x-3|) & = \lim\limits_{x \to 3^-} [2x-(x-3)]\\
\phantom{x} & = \lim\limits_{x \to 3^-} (x+3)\\
\phantom{x} & = 3+3\\
\phantom{x} & = 6
\end{aligned}
\end{equation}
$


The left and right hand limits are equal. Therefore, the limit exists and is equal to 6.