Evaluating a limit by factoring and canceling

Direct substitution at $x=1$ gives $\frac{1-3+2}{1-1} = \frac{0}{0}$, an indeterminate form, so factoring is needed. Factor an $x$ from both numerator and denominator: numerator $x^3-3x^2+2x = x(x^2-3x+2)$, and denominator $x^2-x = x(x-1)$. The quadratic factors further as $x^2-3x+2=(x-1)(x-2)$, so $f(x) = \dfrac{x(x-1)(x-2)}{x(x-1)}$.

For $x \neq 0,1$, the common factors $x$ and $(x-1)$ cancel, leaving $f(x) = x-2$. Since limits depend only on behavior near $x=1$ (not at $x=1$ itself), we can substitute directly into the simplified expression: $\lim_{x\to 1}(x-2) = 1-2 = -1$.

After canceling the common factors of x and x minus one from numerator and denominator, the function simplified to x minus two. This simplified form was valid for all x except zero and one, and allowed direct substitution to evaluate the limit. The answer is x minus two.