Right-hand limit at x = 3

Factor the numerator: $x^2 - 5x + 6 = (x-2)(x-3)$. For $x > 3$, we have $x - 3 > 0$, so $|x-3| = x-3$. Thus for $x > 3$, $f(x) = \frac{(x-2)(x-3)}{x-3} = x-2.$

Since $f(x) = x - 2$ for all $x$ near $3$ from the right, the right-hand limit is found by direct substitution into this simplified expression: $\lim_{x\to 3^+} f(x) = 3 - 2 = 1.$

As before, $x^2 - 5x + 6 = (x-2)(x-3)$. For $x < 3$, we have $x - 3 < 0$, so $|x - 3| = -(x-3) = 3-x$. Thus for $x < 3$, $f(x) = \frac{(x-2)(x-3)}{-(x-3)} = -(x-2) = 2-x.$

Since $f(x) = 2 - x$ for all $x$ near $3$ from the left, direct substitution gives $\lim_{x\to 3^-} f(x) = 2 - 3 = -1.$

From part (a), $\lim_{x\to 3^+} f(x) = 1$, and from part (b), $\lim_{x\to 3^-} f(x) = -1$. These two one-sided limits are not equal to each other.

Since the left-hand and right-hand limits disagree, the two-sided limit $\lim_{x\to 3} f(x)$ does not exist.