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The given limit is:

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

To solve this, we will follow these steps:

### Step 1: Direct Substitution
First, substitute \( x = 3 \) directly into the expression:

\[
\frac{3^2 + 3 - 3}{3^2 - 2(3)} = \frac{9 + 3 - 3}{9 - 6} = \frac{9}{3} = 3
\]

### Final Answer:
The value of the limit is \( 3 \).

Would you like any more details or have any further questions about this process?

Here are 5 related questions you could explore:
1. What happens if direct substitution gives an indeterminate form like \( \frac{0}{0} \)?
2. How can factoring be used to simplify limits involving polynomials?
3. What are other methods to evaluate limits if substitution doesn't work?
4. How can L'Hôpital's Rule help with limits?
5. What are the conditions under which limits exist?

**Tip:** Always check for indeterminate forms when calculating limits, as they may require additional algebraic simplifications or alternative methods!

Find the limit \(\lim_{{x \to 3}} \frac{x^2 + x - 3}{x^2 - 2x}\).

Learn how to find the limit of the function \(\frac{x^2 + x - 3}{x^2 - 2x}\) as x approaches 3 using the direct substitution method. This solution walks through each step and provides insights into key calculus concepts, suitable for Grades 11-12.

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