Question

compute each of the following limits. use exact values.
lim_{a→4^{-}}\frac{|2a - 8|}{a - 4}=2
lim_{a→4^{+}}\frac{|2a - 8|}{a - 4}=
lim_{a→4}\frac{|2a - 8|}{a - 4}=
question help: video

Explanation:

Step1: Analyze left - hand limit

When $a\to4^{-}$, $2a - 8<0$, so $|2a - 8|=-(2a - 8)$. Then $\lim_{a\to4^{-}}\frac{|2a - 8|}{a - 4}=\lim_{a\to4^{-}}\frac{-(2a - 8)}{a - 4}=\lim_{a\to4^{-}}\frac{-2(a - 4)}{a - 4}=- 2$.

Step2: Analyze right - hand limit

When $a\to4^{+}$, $2a - 8>0$, so $|2a - 8|=2a - 8$. Then $\lim_{a\to4^{+}}\frac{|2a - 8|}{a - 4}=\lim_{a\to4^{+}}\frac{2a - 8}{a - 4}=\lim_{a\to4^{+}}\frac{2(a - 4)}{a - 4}=2$.

Step3: Analyze two - sided limit

Since $\lim_{a\to4^{-}}\frac{|2a - 8|}{a - 4}=-2$ and $\lim_{a\to4^{+}}\frac{|2a - 8|}{a - 4}=2$, $\lim_{a\to4}\frac{|2a - 8|}{a - 4}$ does not exist because the left - hand limit and the right - hand limit are not equal.

Answer:

$\lim_{a\to4^{-}}\frac{|2a - 8|}{a - 4}=-2$
$\lim_{a\to4^{+}}\frac{|2a - 8|}{a - 4}=2$
$\lim_{a\to4}\frac{|2a - 8|}{a - 4}$ does not exist