To solve this limit problem:

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

we can follow these steps:

1. Identify the form: Substituting $x = 3$ directly gives: $\frac{\sqrt{3 + 6} - 3}{3^2 - 9} = \frac{3 - 3}{9 - 9} = \frac{0}{0}$ which is an indeterminate form, so we need to simplify the expression.

2. Factor the denominator: Notice that $x^2 - 9$ can be factored as: $x^2 - 9 = (x - 3)(x + 3)$

   So the limit becomes: $\lim_{{x \to 3}} \frac{\sqrt{x + 6} - 3}{(x - 3)(x + 3)}$

3. Rationalize the numerator: To handle the square root in the numerator, multiply the numerator and the denominator by the conjugate of the numerator, $\sqrt{x + 6} + 3$: $\lim_{{x \to 3}} \frac{(\sqrt{x + 6} - 3)(\sqrt{x + 6} + 3)}{(x - 3)(x + 3)(\sqrt{x + 6} + 3)}$

4. Simplify the expression: Expanding the numerator using the difference of squares gives: $(\sqrt{x + 6} - 3)(\sqrt{x + 6} + 3) = (x + 6) - 9 = x - 3$

   So the expression simplifies to: $\lim_{{x \to 3}} \frac{x - 3}{(x - 3)(x + 3)(\sqrt{x + 6} + 3)}$

5. Cancel terms: Now, $x - 3$ can be canceled from the numerator and the denominator: $\lim_{{x \to 3}} \frac{1}{(x + 3)(\sqrt{x + 6} + 3)}$

6. Evaluate the limit: Substitute $x = 3$: $\frac{1}{(3 + 3)(\sqrt{3 + 6} + 3)} = \frac{1}{6 \cdot (3 + 3)} = \frac{1}{36}$

Answer: The limit is $\frac{1}{36}$.

Would you like further clarification on any steps?

Here are some related questions for practice:

1. How do you rationalize the numerator in a limit problem involving square roots?
2. What is the difference of squares formula, and how is it useful in simplifying limits?
3. How do you handle indeterminate forms in limits?
4. What other methods, besides rationalization, can help solve limits involving radicals?
5. How can factoring help in solving limits?

Tip: Always look for factorizable expressions or common terms that can be canceled to simplify a limit expression.