## Calculus 8th Edition

Theorem 1 $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if and only if $\displaystyle \lim_{x\rightarrow a^{-}}f(x)=L=\lim_{x\rightarrow a^{+}}f(x)$ --------- $|x+6|=\left\{\begin{array}{lll} x+6 & if & x \geq -6\\ -(x+6) & if & x < -6 \end{array}\right.$ Approaching x=-6 from the right, $\displaystyle \lim_{x\rightarrow-6+}\frac{2x+12}{|x+6|}=\lim_{x\rightarrow-6+}\frac{2(x+6)}{x+6}=2$ Approaching x=-6 from the left, $\displaystyle \lim_{x\rightarrow-6^{-}}\frac{2x+12}{|x+6|}=\lim_{x\rightarrow-6^{-}}\frac{2(x+6)}{-(x+6)}=-2$ The left and right limits exist, but are NOT EQUAL, so by Th. 1, $\displaystyle \lim_{x\rightarrow-6}\frac{2x+12}{|x+6|}$ does not exist.