Evaluating a limit by factoring and canceling

Substituting $x=1$ directly gives $\frac{1-4+3}{1-1} = \frac{0}{0}$, an indeterminate form, so factoring is needed. The numerator factors as $x^2 - 4x + 3 = (x-1)(x-3)$, and the denominator factors as $x^3 - x = x(x^2-1) = x(x-1)(x+1)$.

Since $x \to 1$ but $x \neq 1$, we can cancel the common factor $(x-1)$ from numerator and denominator, leaving $\frac{x-3}{x(x+1)}$. Substituting $x = 1$ into this simplified expression gives $\frac{1-3}{1(2)} = \frac{-2}{2} = -1$.