Classify the discontinuity at x = 2

Factor both the numerator and denominator: $x^2 - 5x + 6 = (x-2)(x-3)$ and $x^2 - 4 = (x-2)(x+2)$. So $w(x) = \dfrac{(x-2)(x-3)}{(x-2)(x+2)}$. Since $x = 2$ makes both factors zero, $w(2)$ is undefined (0/0 form), but the factor $(x-2)$ cancels for $x \neq 2$.

For $x \neq 2$, $w(x) = \dfrac{x-3}{x+2}$, so $\lim_{x\to 2} w(x) = \dfrac{2-3}{2+2} = -\dfrac{1}{4}$. This limit exists and is finite, but $f(2)$ is undefined, so the second and third conditions of continuity fail while the limit itself exists. This means the graph has a hole at $(2, -1/4)$ that could be patched by defining $w(2) = -1/4$.