To find the limit:

$\lim_{x \to -2} \frac{x^2 + 8x + 12}{x + 2}$

Let's first try to substitute $x = -2$ directly into the expression. Doing so gives:

$\frac{(-2)^2 + 8(-2) + 12}{-2 + 2} = \frac{4 - 16 + 12}{0} = \frac{0}{0}$

This results in an indeterminate form $\frac{0}{0}$, which means we need to simplify the expression further, typically by factoring.

Step 1: Factor the numerator

We can factor the quadratic expression $x^2 + 8x + 12$:

$x^2 + 8x + 12 = (x + 2)(x + 6)$

Step 2: Simplify the expression

Now, substitute the factored form into the limit expression:

$\frac{(x + 2)(x + 6)}{x + 2}$

We can cancel out the common factor $(x + 2)$ in the numerator and denominator:

$= x + 6 \quad \text{for} \quad x \neq -2$

Step 3: Take the limit

Now that the expression is simplified, we can substitute $x = -2$:

$\lim_{x \to -2} (x + 6) = -2 + 6 = 4$

Thus, the limit is:

$\boxed{4}$

Would you like more details on any step, or do you have any questions?

Here are 5 related questions for further practice:

1. Find the limit $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.
2. Determine the limit $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$.
3. Evaluate the limit $\lim_{x \to 0} \frac{x^3 + 3x^2}{x}$.
4. Solve the limit $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$.
5. Find the limit $\lim_{x \to -1} \frac{x^3 + 1}{x + 1}$.

Tip: When you get a $\frac{0}{0}$ indeterminate form, factoring or applying L'Hopital's Rule are common methods to resolve it.