To evaluate the limit

$\lim_{x \to 4} \frac{\sqrt{x+5} - 3}{x^2 - x - 12}$

we first try direct substitution of $x = 4$:

$\text{Numerator: } \sqrt{4+5} - 3 = \sqrt{9} - 3 = 3 - 3 = 0$

$\text{Denominator: } 4^2 - 4 - 12 = 16 - 4 - 12 = 0$

Since both the numerator and denominator result in 0, we have an indeterminate form $\frac{0}{0}$. To resolve this, we can apply L'Hôpital's Rule, but before doing so, let's try factoring or simplifying.

Step 1: Simplifying the Denominator

The denominator can be factored as follows:

$x^2 - x - 12 = (x - 4)(x + 3)$

So the limit becomes:

$\lim_{x \to 4} \frac{\sqrt{x+5} - 3}{(x - 4)(x + 3)}$

Step 2: Rationalizing the Numerator

We can try rationalizing the numerator by multiplying the numerator and denominator by the conjugate of the numerator:

$\frac{\sqrt{x+5} - 3}{(x - 4)(x + 3)} \cdot \frac{\sqrt{x+5} + 3}{\sqrt{x+5} + 3} = \frac{(\sqrt{x+5})^2 - 3^2}{(x - 4)(x + 3)(\sqrt{x+5} + 3)}$

Simplifying the numerator:

$(\sqrt{x+5})^2 - 3^2 = (x+5) - 9 = x - 4$

So the limit simplifies to:

$\lim_{x \to 4} \frac{x - 4}{(x - 4)(x + 3)(\sqrt{x+5} + 3)} = \lim_{x \to 4} \frac{1}{(x + 3)(\sqrt{x+5} + 3)}$

Step 3: Direct Substitution

Now, we can substitute $x = 4$ into the simplified expression:

$\frac{1}{(4 + 3)(\sqrt{4 + 5} + 3)} = \frac{1}{7 \times (3 + 3)} = \frac{1}{7 \times 6} = \frac{1}{42}$

Thus, the limit is:

$\boxed{\frac{1}{42}}$

Would you like any further details or explanations? Here are some related questions:

1. How do you apply L'Hôpital's Rule to indeterminate forms?
2. What is the conjugate of a binomial, and how is it used in limits?
3. How do you factor quadratic expressions like $x^2 - x - 12$?
4. When is it appropriate to rationalize the numerator in a limit problem?
5. What are other techniques to solve indeterminate forms in limits?

Tip: Rationalizing the numerator is a useful strategy when dealing with square roots in limits.