Question

find the limit, if it exists. (if an answer does not exist, enter dne.)
\\(lim_{x
ightarrow - 3}\frac{|x + 3|}{4x + 12}\\)

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Explanation:

Step1: Factor the denominator

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

Step2: Consider left - hand and right - hand limits

Left - hand limit ($x

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

Right - hand limit ($x

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

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 $\lim_{x
ightarrow - 3}\frac{|x + 3|}{4(x + 3)}$ does not exist.

Answer:

DNE