find the following limit limit->-2 (x^2+8x+12)/(x+2)

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.

Learn how to evaluate the limit of (x^2 + 8x + 12) / (x + 2) as x approaches -2. This step-by-step solution explains factoring a quadratic expression and simplifying indeterminate forms, suitable for students in Grades 10-12.

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