Answer
$\lim\limits_{x\to3}(2x+|x-3|)=6$
Work Step by Step
$\lim\limits_{x\to3}(2x+|x-3|)$
$=\lim\limits_{x\to3}2x+\lim\limits_{x\to3}|x-3|$
$=6+\lim\limits_{x\to3}|x-3|$
We know that
$|x-3|=$$\left \{
\begin{array} {c l}
x-3 && x\geq3\\
-(x-3)=3-x && x<3
\end{array}
\right.$
Therefore,
$\lim\limits_{x\to3^+}|x-3|=\lim\limits_{x\to3^+}(x-3)=3-3=0$
$\lim\limits_{x\to3^-}|x-3|=\lim\limits_{x\to3^-}(3-x)=3-3=0$
We see that $\lim\limits_{x\to3^+}|x-3|=\lim\limits_{x\to3^-}|x-3|=0$
So, $\lim\limits_{x\to3}|x-3|=0$
Therefore, $$\lim\limits_{x\to3}(2x+|x-3|)=6+0=6$$
