Question

use numerical or graphical evidence to determine the left and right hand limits of the $lim_{x
ightarrow4^{-}}\frac{|4x - 16|}{4 - x}=$ $lim_{x
ightarrow4^{+}}\frac{|4x - 16|}{4 - x}=$ question help: message instructor

Explanation:

Step1: Analyze the absolute - value function

First, rewrite \(|4x - 16|=|4(x - 4)| = 4|x - 4|\).

Step2: Calculate the left - hand limit (\(x\to4^{-}\))

When \(x\to4^{-}\), \(x-4\lt0\), so \(|x - 4|=-(x - 4)\). Then \(\lim_{x\to4^{-}}\frac{|4x - 16|}{4 - x}=\lim_{x\to4^{-}}\frac{4|x - 4|}{4 - x}=\lim_{x\to4^{-}}\frac{4-(x - 4)}{4 - x}\). Simplify the expression: \(\lim_{x\to4^{-}}\frac{-4(x - 4)}{4 - x}=\lim_{x\to4^{-}}\frac{-4(x - 4)}{-(x - 4)} = 4\).

Step3: Calculate the right - hand limit (\(x\to4^{+}\))

When \(x\to4^{+}\), \(x - 4\gt0\), so \(|x - 4|=x - 4\). Then \(\lim_{x\to4^{+}}\frac{|4x - 16|}{4 - x}=\lim_{x\to4^{+}}\frac{4|x - 4|}{4 - x}=\lim_{x\to4^{+}}\frac{4(x - 4)}{4 - x}\). Since \(x-4\) and \(4 - x\) are negatives of each other, \(\lim_{x\to4^{+}}\frac{4(x - 4)}{4 - x}=-4\).

Answer:

\(\lim_{x\to4^{-}}\frac{|4x - 16|}{4 - x}=4\)
\(\lim_{x\to4^{+}}\frac{|4x - 16|}{4 - x}=-4\)