Classify the discontinuity at x = 4

For $x < 4$, we have $x - 4 < 0$, so $|x-4| = -(x-4)$. Thus $q(x) = \dfrac{-(x-4)}{x-4} = -1$ for all $x < 4$, which gives $\lim_{x \to 4^-} q(x) = -1$.

For $x > 4$, we have $x - 4 > 0$, so $|x-4| = x-4$. Thus $q(x) = \dfrac{x-4}{x-4} = 1$ for all $x > 4$, which gives $\lim_{x \to 4^+} q(x) = 1$. Since both one-sided limits exist but are unequal ($-1 \ne 1$), the two-sided limit does not exist, and the discontinuity at $x=4$ is a jump discontinuity.