Question

this question has several parts that must be completed sequentially. if you skip a part of the question, you will not receive any points. tutorial exercise. find the limit, if it exists. $lim_{x
ightarrow - 4}\frac{3x + 12}{|x + 4|}$. part 1 of 5. recall that $|x + 4|=\begin{cases}x + 4&xgeq - 4\\-(x + 4)&xlt - 4end{cases}$. 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 $lim_{x
ightarrow - 4^{-}}\frac{3x + 12}{|x + 4|}=lim_{x
ightarrow - 4^{-}}\frac{3x + 12}{square}$. submit. skip (you cannot come back). resources. read it

Explanation:

Step1: Analyze the absolute - value for left - hand limit

When $x\to - 4^{-}$, $x<-4$. So, $|x + 4|=-(x + 4)$.

Step2: Substitute the absolute - value expression

We have $\lim_{x\to - 4^{-}}\frac{3x + 12}{|x + 4|}=\lim_{x\to - 4^{-}}\frac{3x + 12}{-(x + 4)}$.
Factor the numerator: $3x+12 = 3(x + 4)$. Then the limit becomes $\lim_{x\to - 4^{-}}\frac{3(x + 4)}{-(x + 4)}$.

Step3: Simplify the expression

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

Answer:

$-3$