Answer
6
Work Step by Step
Theorem 1
$\displaystyle \lim_{x\rightarrow a}f(x)=L$ if and only if $\displaystyle \lim_{x\rightarrow a^{-}}f(x)=L=\lim_{x\rightarrow a^{+}}f(x)$
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$|x-3|=\left\{\begin{array}{lll}
x-3 & if & x \geq 3\\
3-x & if & x < 3
\end{array}\right.$
so $f(x)=\left\{\begin{array}{lll}
2x+x-3 & if & x \geq 3\\
2x+3-x & if & x < 3
\end{array}\right.$
Approaching x=3 from the right,
$\displaystyle \lim_{x\rightarrow 3+}f(x)=\lim_{x\rightarrow 3+}(2x+x-3)$
$=\displaystyle \lim_{x\rightarrow 3+}(3x-3)=3(3)-3=6$
Approaching x=3 from the left,
$\displaystyle \lim_{x\rightarrow 3^{-}}f(x)=\lim_{x\rightarrow \mathrm{s}-}(2x+3-x)$
$=\displaystyle \lim_{x\rightarrow 3^{-}}(x+3)=3+3=6$.
The left and right limits exist and are equal, so
$\displaystyle \lim_{x\rightarrow 3}(2x+|x-3|)=6$.
