The given limit is:

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

To evaluate this limit, let's use algebraic manipulation by multiplying the numerator and denominator by the conjugate of the numerator to eliminate the square root.

1. Multiply the expression by $\frac{\sqrt{x + 3} + 3}{\sqrt{x + 3} + 3}$:

$\frac{\sqrt{x + 3} - 3}{x - 6} \cdot \frac{\sqrt{x + 3} + 3}{\sqrt{x + 3} + 3} = \frac{(\sqrt{x + 3} - 3)(\sqrt{x + 3} + 3)}{(x - 6)(\sqrt{x + 3} + 3)}$

2. Simplify the numerator using the difference of squares:

$(\sqrt{x + 3} - 3)(\sqrt{x + 3} + 3) = (x + 3) - 9 = x - 6$

Now, the expression simplifies to:

$\frac{x - 6}{(x - 6)(\sqrt{x + 3} + 3)}$

3. Cancel the common factor $(x - 6)$:

$\frac{1}{\sqrt{x + 3} + 3}$

4. Substitute $x = 6$:

$\frac{1}{\sqrt{6 + 3} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Therefore, the limit is:

$\boxed{\frac{1}{6}}$

Would you like further details or have any questions? Here are some related problems to explore:

1. What happens if the limit is taken as $x \to 0$ instead of $x \to 6$?
2. How can you evaluate limits involving other types of indeterminate forms?
3. How would the result change if the expression were $\sqrt{x - 3}$ instead of $\sqrt{x + 3}$?
4. What are other methods to simplify limits involving square roots?
5. How can L'Hopital's Rule be applied to this problem?

Tip: Multiplying by the conjugate is a common technique to simplify expressions with square roots when direct substitution leads to an indeterminate form.