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The given limit expression is:

\[
\lim_{{x \to 1}} \frac{{x^2 - x}}{{x^2 + 2x - 3}}
\]

### Step-by-Step Solution:

1. **Factor the denominator:**

   The denominator is \(x^2 + 2x - 3\). We can factor it as:
   \[
   x^2 + 2x - 3 = (x - 1)(x + 3)
   \]

2. **Factor the numerator:**

   The numerator is \(x^2 - x\). We can factor this as:
   \[
   x^2 - x = x(x - 1)
   \]

3. **Substitute the factored form:**

   Substituting the factored forms of both the numerator and denominator, we get:
   \[
   \lim_{{x \to 1}} \frac{{x(x - 1)}}{{(x - 1)(x + 3)}}
   \]

4. **Cancel common terms:**

   The terms \((x - 1)\) in the numerator and denominator can be canceled (since \(x \to 1\), \((x - 1)\) is not zero in the limit):
   \[
   \lim_{{x \to 1}} \frac{x}{{x + 3}}
   \]

5. **Evaluate the limit:**

   Now, substitute \(x = 1\):
   \[
   \frac{1}{{1 + 3}} = \frac{1}{4}
   \]

### Final Answer:

\[
\lim_{{x \to 1}} \frac{{x^2 - x}}{{x^2 + 2x - 3}} = \frac{1}{4}
\]

Would you like further details or clarification on any step?

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Here are five related questions for further practice:

1. What is the limit \(\lim_{{x \to 2}} \frac{{x^2 - 4}}{{x^2 - 5x + 6}}\)?
2. How do you evaluate limits using L'Hopital's Rule?
3. Can all limits be evaluated by factoring?
4. What happens if a limit results in \(\frac{0}{0}\)?
5. How would you find the limit at infinity for a rational function?

**Tip:** Always check if the expression can be simplified or factored before applying more advanced techniques like L'Hopital's Rule.

lim(x → 1) [(x² - x) / (x² + 2x - 3)]

Explore the step-by-step solution to the limit problem lim(x → 1) [(x² - x) / (x² + 2x - 3)] by factoring the numerator and denominator. This guide walks through simplifying and evaluating limits, suitable for high school students in Grades 10-12.

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