To solve the limit of the function as $x \to -2$:

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

Step-by-step solution:

1. Substitute $x = -2$ directly into the function: $\frac{(-2)^2 + (-2) - 2}{-2 + 2} = \frac{4 - 2 - 2}{0} = \frac{0}{0}$ This results in an indeterminate form, $\frac{0}{0}$, so we need to simplify the function.

2. Factor the numerator: $x^2 + x - 2 = (x + 2)(x - 1)$ So the expression becomes: $\frac{(x + 2)(x - 1)}{x + 2}$

3. Cancel the common factor $(x + 2)$ from the numerator and denominator: $\frac{(x + 2)(x - 1)}{x + 2} = x - 1, \quad \text{for} \, x \neq -2$

4. Take the limit of the simplified expression: $\lim_{x \to -2} (x - 1) = -2 - 1 = -3$

Thus, the limit is: $\boxed{-3}$

Would you like more details or have any questions?

Related questions:

1. What if the function had more complicated factors in the numerator?
2. How do you handle limits that result in an infinity form?
3. What happens if the function isn’t factorable in limits like this?
4. Can L'Hopital's Rule be applied here instead of factoring?
5. What if $x \to 2$ instead of $x \to -2$?

Tip:

When faced with indeterminate forms like $\frac{0}{0}$, factorization or using L'Hopital's Rule are good strategies to simplify the expression.