The given limit problem is:

$\lim_{{x \to 3}} \frac{x^2 - 9}{x - 3}$

Solution:

1. Factor the numerator: The numerator $x^2 - 9$ is a difference of squares, so it can be factored as: $x^2 - 9 = (x - 3)(x + 3)$

2. Rewrite the expression: Now, the expression becomes: $\frac{(x - 3)(x + 3)}{x - 3}$

3. Simplify: Since $x - 3$ appears in both the numerator and the denominator, and $x \neq 3$ (as we are finding the limit), they cancel out: $\frac{(x - 3)(x + 3)}{x - 3} = x + 3$

4. Evaluate the limit: Now that we have simplified the expression, we can directly substitute $x = 3$ into $x + 3$: $\lim_{{x \to 3}} (x + 3) = 3 + 3 = 6$

Thus, the solution is:

$\lim_{{x \to 3}} \frac{x^2 - 9}{x - 3} = 6$

Would you like more details on any part of this solution or have additional questions?

Here are five related questions to expand on this:

1. What happens if you don't factor the numerator in a limit problem like this?
2. How do we handle limits that result in $0/0$ indeterminate forms?
3. What is L'Hôpital's rule, and how could it apply to this type of problem?
4. How do we handle limits approaching infinity for rational functions?
5. Can limits always be simplified by factoring, or are there other techniques?

Tip: Always check for common factors in the numerator and denominator when faced with a limit problem involving polynomials. It can help simplify the expression before taking the limit.