Question

part 1 of 5 recall that $|x + 4|=\begin{cases}x + 4&xgeq - 4\\-(x + 4)&x < - 4end{cases}$. therefore, we will need to check the limits when approaching from the left and from the right. we will start by checking the limit when approaching from the left. as $x$ approaches $-4$ from the left, we have $lim_{x
ightarrow - 4^{-}}\frac{3x + 12}{|x + 4|}=lim_{x
ightarrow - 4^{-}}\frac{3x + 12}{square}$.

Explanation:

Step1: Factor the numerator

Factor \(3x + 12\) as \(3(x + 4)\). So we have \(\lim_{x
ightarrow - 4^{-}}\frac{3x + 12}{|x + 4|}=\lim_{x
ightarrow - 4^{-}}\frac{3(x + 4)}{|x + 4|}\).

Step2: Determine the value of \(|x + 4|\) for \(x

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

Step3: Simplify the limit expression

Cancel out the \((x + 4)\) terms (since \(x
eq - 4\) when taking the limit). We get \(\lim_{x
ightarrow - 4^{-}}\frac{3(x + 4)}{-(x + 4)}=\lim_{x
ightarrow - 4^{-}}-3=-3\).

Answer:

\(-3\)