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?

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.