Question

find the limit, if it exists. (if an answer does not exist, enter dne.)
lim_{x
ightarrow - 5}\frac{8x + 40}{|x + 5|}

Explanation:

Step1: Factor the numerator

Factor out 8 from the numerator $8x + 40$, we get $8(x + 5)$. So the function becomes $\lim_{x
ightarrow - 5}\frac{8(x + 5)}{|x + 5|}$.

Step2: Consider left - hand and right - hand limits

Left - hand limit ($x

ightarrow - 5^{-}$):
When $x
ightarrow - 5^{-}$, $x+5<0$, then $|x + 5|=-(x + 5)$. So $\lim_{x
ightarrow - 5^{-}}\frac{8(x + 5)}{|x + 5|}=\lim_{x
ightarrow - 5^{-}}\frac{8(x + 5)}{-(x + 5)}=-8$.

Right - hand limit ($x

ightarrow - 5^{+}$):
When $x
ightarrow - 5^{+}$, $x + 5>0$, then $|x + 5|=x + 5$. So $\lim_{x
ightarrow - 5^{+}}\frac{8(x + 5)}{|x + 5|}=\lim_{x
ightarrow - 5^{+}}\frac{8(x + 5)}{x + 5}=8$.

Step3: Determine the limit

Since the left - hand limit $\lim_{x
ightarrow - 5^{-}}\frac{8x + 40}{|x + 5|}=-8$ and the right - hand limit $\lim_{x
ightarrow - 5^{+}}\frac{8x + 40}{|x + 5|}=8$, and $-8
eq8$, the two - sided limit $\lim_{x
ightarrow - 5}\frac{8x + 40}{|x + 5|}$ does not exist.

Answer:

DNE