The problem presented is to evaluate the limit:

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

Step 1: Factor the numerator

First, we notice that the numerator $x^2 - 9$ is a difference of squares, which can be factored as follows:

$x^2 - 9 = (x - 3)(x + 3)$

So the expression becomes:

$\lim_{{x \to 3}} \frac{(x - 3)(x + 3)}{x - 3}$

Step 2: Simplify the expression

Since $x - 3$ is a common factor in both the numerator and denominator, we can cancel it out (as long as $x$ is not equal to 3, which it isn't in the context of taking a limit):

$\lim_{{x \to 3}} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{{x \to 3}} (x + 3)$

Step 3: Evaluate the limit

Now, substitute $x = 3$ into the simplified expression:

$\lim_{{x \to 3}} (x + 3) = 3 + 3 = 6$

Final Answer

The limit is:

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

Would you like further details or have any questions?

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Related Questions:

1. What is the limit of $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$?
2. How do you evaluate $\lim_{{x \to 0}} \frac{x^2 - 1}{x}$?
3. Can you explain the difference between a finite limit and an infinite limit?
4. What happens if the numerator is not factorable?
5. How does L'Hôpital's Rule apply to limits of the form $\frac{0}{0}$?

Tip: When solving limits that result in an indeterminate form like $\frac{0}{0}$, factoring is often the first step to simplify the expression.