Question

e) $lim_{x
ightarrow - 4}\frac{|x + 4|}{2x+8}$

Explanation:

Step1: Factor the denominator

Factor \(2x + 8\) as \(2(x + 4)\). So the limit becomes \(\lim_{x
ightarrow - 4}\frac{|x + 4|}{2(x + 4)}\).

Step2: Consider left - hand and right - hand limits

Left - hand limit (\(x

ightarrow - 4^{-}\)):
When \(x
ightarrow - 4^{-}\), \(x+4<0\), so \(|x + 4|=-(x + 4)\). Then \(\lim_{x
ightarrow - 4^{-}}\frac{|x + 4|}{2(x + 4)}=\lim_{x
ightarrow - 4^{-}}\frac{-(x + 4)}{2(x + 4)}=-\frac{1}{2}\).

Right - hand limit (\(x

ightarrow - 4^{+}\)):
When \(x
ightarrow - 4^{+}\), \(x + 4>0\), so \(|x + 4|=x + 4\). Then \(\lim_{x
ightarrow - 4^{+}}\frac{|x + 4|}{2(x + 4)}=\lim_{x
ightarrow - 4^{+}}\frac{x + 4}{2(x + 4)}=\frac{1}{2}\).
Since the left - hand limit \(-\frac{1}{2}
eq\frac{1}{2}\) (the right - hand limit), the limit \(\lim_{x
ightarrow - 4}\frac{|x + 4|}{2(x + 4)}\) does not exist.

Answer:

The limit does not exist.