Question

use numerical or graphical evidence to determine the left and right hand limits of the function. \\(\lim_{x\to 3^-} \frac{|24 - 8x|}{3 - x} = \\) \\(\lim_{x\to 3^+} \frac{|24 - 8x|}{3 - x} = \\)

Explanation:

Step1: Simplify absolute value for left limit

For $x \to 3^-$, $x < 3$, so $8x < 24$, $24-8x > 0$. Thus $|24-8x|=24-8x$.
$\lim_{x \to 3^-} \frac{24-8x}{3-x} = \lim_{x \to 3^-} \frac{8(3-x)}{3-x}$

Step2: Cancel common terms (left limit)

Cancel $(3-x)$ (non-zero as $x
eq 3$).
$\lim_{x \to 3^-} 8 = 8$

Step3: Simplify absolute value for right limit

For $x \to 3^+$, $x > 3$, so $8x > 24$, $24-8x < 0$. Thus $|24-8x|=-(24-8x)=8x-24$.
$\lim_{x \to 3^+} \frac{8x-24}{3-x} = \lim_{x \to 3^+} \frac{8(x-3)}{3-x}$

Step4: Rewrite and cancel terms (right limit)

Rewrite $(x-3)=-(3-x)$, then cancel $(3-x)$.
$\lim_{x \to 3^+} \frac{8(-(3-x))}{3-x} = \lim_{x \to 3^+} -8 = -8$

Answer:

$\lim_{x \to 3^-} \frac{|24-8x|}{3-x} = 8$
$\lim_{x \to 3^+} \frac{|24-8x|}{3-x} = -8$