Let's analyze the limit: $\lim_{x \to -2} \frac{x^2 - 4x - 12}{x^2 + 3x + 2}.$

Step 1: Substitute $x = -2$

To check if we can directly substitute $x = -2$: $\frac{(-2)^2 - 4(-2) - 12}{(-2)^2 + 3(-2) + 2} = \frac{4 + 8 - 12}{4 - 6 + 2} = \frac{0}{0}.$ This results in an indeterminate form $\frac{0}{0}$, so we need to simplify the expression.

Step 2: Factor the numerator and denominator

Factor both the numerator and the denominator:

• Numerator: $x^2 - 4x - 12 = (x - 6)(x + 2)$.
• Denominator: $x^2 + 3x + 2 = (x + 2)(x + 1)$.

The expression now looks like this: $\lim_{x \to -2} \frac{(x - 6)(x + 2)}{(x + 2)(x + 1)}.$

Step 3: Cancel the common factor

We can cancel the $(x + 2)$ term (since it's not zero when $x \neq -2$): $\lim_{x \to -2} \frac{x - 6}{x + 1}.$

Step 4: Substitute $x = -2$ again

Now, substitute $x = -2$ into the simplified expression: $\frac{-2 - 6}{-2 + 1} = \frac{-8}{-1} = 8.$

Conclusion

The limit is: $\lim_{x \to -2} \frac{x^2 - 4x - 12}{x^2 + 3x + 2} = 8.$

Would you like more details or have any questions about this solution?

Related Questions

1. What happens when the limit results in an indeterminate form like $\frac{0}{0}$?
2. How do you factor quadratic expressions like $x^2 - 4x - 12$?
3. What is the meaning of canceling terms in a rational expression?
4. How do limits work when approaching infinity?
5. What other methods can be used to solve indeterminate forms besides factoring?

Tip

Always check for indeterminate forms first when dealing with limits to decide the best method for simplification.