Question

1. (10 points) evaluate \\(\lim\limits_{x \to 3} \frac{|x - 3|}{x - 3}\\).

Explanation:

Step1: Split absolute value by cases

For $x > 3$: $|x-3|=x-3$; for $x < 3$: $|x-3|=-(x-3)$

Step2: Compute right-hand limit

$\lim_{x \to 3^+} \frac{x-3}{x-3} = \lim_{x \to 3^+} 1 = 1$

Step3: Compute left-hand limit

$\lim_{x \to 3^-} \frac{-(x-3)}{x-3} = \lim_{x \to 3^-} (-1) = -1$

Step4: Compare two limits

Right-hand limit ≠ Left-hand limit

Answer:

The limit does not exist.