## Calculus 8th Edition

Theorem 1 $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if and only if $\displaystyle \lim_{x\rightarrow a^{-}}f(x)=L=\lim_{x\rightarrow a^{+}}f(x)$ --------- $|x-3|=\left\{\begin{array}{lll} x-3 & if & x \geq 3\\ 3-x & if & x < 3 \end{array}\right.$ 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 $\displaystyle \lim_{x\rightarrow 3}(2x+|x-3|)=6$.