5) $lim_{x\rightarrow - 4}\frac{3x + 12}{|x + 4|}$

Explanation:

Step1: Evaluate the left-hand limit.

As $x \to -4^-$, $x < -4$, so $x+4 < 0$. Thus, $|x+4| = -(x+4)$. $$ \lim_{x \to -4^-} \frac{3x+12}{|x+4|} = \lim_{x \to -4^-} \frac{3(x+4)}{-(x+4)} $$ $$ = \lim_{x \to -4^-} -3 = -3 $$

Step2: Evaluate the right-hand limit.

As $x \to -4^+$, $x > -4$, so $x+4 > 0$. Thus, $|x+4| = x+4$. $$ \lim_{x \to -4^+} \frac{3x+12}{|x+4|} = \lim_{x \to -4^+} \frac{3(x+4)}{x+4} $$ $$ = \lim_{x \to -4^+} 3 = 3 $$

Step3: Compare the left-hand and right-hand limits.

Since the left-hand limit ($-3$) is not equal to the right-hand limit ($3$), the limit does not exist. $$ \lim_{x \to -4^-} f(x) \neq \lim_{x \to -4^+} f(x) $$

Answer:

The limit does not exist.