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.