The problem asks us to compute the limit:

$\lim_{x \to 1} \frac{x^2 - 3x + 2}{x^2 - x}.$

Step-by-step solution:

1. Factorize the numerator and denominator:

   • The numerator $x^2 - 3x + 2$ can be factorized as: $x^2 - 3x + 2 = (x - 1)(x - 2).$
   • The denominator $x^2 - x$ can be factorized as: $x^2 - x = x(x - 1).$

2. Rewrite the expression: Substitute the factored forms into the limit: $\lim_{x \to 1} \frac{(x - 1)(x - 2)}{x(x - 1)}.$

3. Simplify the expression: Cancel out the common factor $(x - 1)$ (valid since $x \neq 1$): $\lim_{x \to 1} \frac{x - 2}{x}.$

4. Evaluate the limit: Substitute $x = 1$ into the simplified expression: $\frac{x - 2}{x} = \frac{1 - 2}{1} = -1.$

Final Answer:

$\lim_{x \to 1} \frac{x^2 - 3x + 2}{x^2 - x} = -1.$

Would you like more details or have any questions? Here are some related problems to expand your knowledge:

1. What happens if the limit is undefined or involves division by zero?
2. How does factoring help in evaluating limits with indeterminate forms like $\frac{0}{0}$?
3. Can you apply L'Hôpital's Rule to this problem? Why or why not?
4. What is the importance of simplifying rational expressions before evaluating limits?
5. What are other techniques to handle $\lim_{x \to a}$ problems when direct substitution fails?

Tip: Always check for common factors in both the numerator and denominator when evaluating limits involving rational functions.