Evaluating a limit by factoring

Direct substitution at $x=2$ gives $\frac{4-10+6}{4-4} = \frac{0}{0}$, an indeterminate form, so factoring is needed. The numerator factors as $x^2 - 5x + 6 = (x-2)(x-3)$, and the denominator factors as $x^2 - 4 = (x-2)(x+2)$.

Since $x \neq 2$ near the point of interest, the common factor $(x-2)$ cancels, leaving $\frac{x-3}{x+2}$. Substituting $x = 2$ into this simplified expression gives $\frac{2-3}{2+2} = \frac{-1}{4}$.

The factor canceled was x minus two. Both the numerator and denominator shared this common factor after factoring, since the numerator factored to x minus two times x minus three, and the denominator factored to x minus two times x plus two. Canceling it removed the zero over zero indeterminate form and left a simplified expression valid for x near but not equal to two.