To evaluate the limit $\lim_{{x \to -3}} \frac{x^2 - x - 12}{x + 3}$, let's proceed with the following steps:

1. Substitute $x = -3$ directly into the expression to see if we can find the limit without further simplification.

$\frac{(-3)^2 - (-3) - 12}{-3 + 3} = \frac{9 + 3 - 12}{0} = \frac{0}{0}$

Since we get an indeterminate form $\frac{0}{0}$, we need to simplify the expression.

2. Factor the numerator:

The numerator $x^2 - x - 12$ can be factored as:

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

3. Simplify the expression:

Now, the limit becomes:

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

We can cancel out the $(x + 3)$ term in the numerator and the denominator (as long as $x \neq -3$):

$\lim_{{x \to -3}} (x - 4)$

4. Evaluate the limit:

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

$-3 - 4 = -7$

Thus, the limit is:

$\boxed{-7}$

Would you like any further details or have any questions?

Here are some related questions:

1. How do you handle limits that result in other indeterminate forms like $\frac{\infty}{\infty}$?
2. What is L'Hôpital's Rule, and when is it used?
3. How do you find the limit of a trigonometric function as $x$ approaches a specific value?
4. What are the methods for factoring quadratic expressions like the one in this problem?
5. How do you evaluate limits at infinity?

Tip: When you encounter an indeterminate form, simplifying the expression by factoring or rationalizing is often the key to finding the limit.