find the limit, if it exists. (if an answer does not exist, enter dne.)\n\\(lim_{x\rightarrow - 3}\frac{|x +…

Explanation:

Step1: Factor the denominator

Factor $4x + 12$ as $4(x + 3)$. So the limit becomes $\lim_{x\rightarrow - 3}\frac{|x + 3|}{4(x + 3)}$.

Step2: Consider left - hand and right - hand limits

Left - hand limit ($x\rightarrow - 3^{-}$):

When $x\rightarrow - 3^{-}$, $x+3<0$, so $|x + 3|=-(x + 3)$. Then $\lim_{x\rightarrow - 3^{-}}\frac{|x + 3|}{4(x + 3)}=\lim_{x\rightarrow - 3^{-}}\frac{-(x + 3)}{4(x + 3)}=-\frac{1}{4}$.

Right - hand limit ($x\rightarrow - 3^{+}$):

When $x\rightarrow - 3^{+}$, $x + 3>0$, so $|x + 3|=x + 3$. Then $\lim_{x\rightarrow - 3^{+}}\frac{|x + 3|}{4(x + 3)}=\lim_{x\rightarrow - 3^{+}}\frac{x + 3}{4(x + 3)}=\frac{1}{4}$.

Step3: Determine the limit

Since the left - hand limit $-\frac{1}{4}$ is not equal to the right - hand limit $\frac{1}{4}$, the limit $\lim_{x\rightarrow - 3}\frac{|x + 3|}{4(x + 3)}$ does not exist.

Answer:

DNE