To evaluate the limit:

$\lim_{x \to -2} \frac{x+2}{x^2 + x - 6}$

Step 1: Factor the denominator

First, let's factor the quadratic expression in the denominator:

$x^2 + x - 6 = (x + 3)(x - 2)$

So the expression becomes:

$\frac{x + 2}{(x + 3)(x - 2)}$

Step 2: Direct substitution

Now, substitute $x = -2$ directly into the expression:

$\frac{-2 + 2}{(-2 + 3)(-2 - 2)} = \frac{0}{(1)(-4)} = 0$

Since the numerator becomes 0 and the denominator is not 0, the limit is:

$\boxed{0}$

Conclusion

The limit as $x \to -2$ of $\frac{x+2}{x^2 + x - 6}$ is 0.

Do you want more details on this calculation or have any other questions?

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Here are 5 related questions:

1. What if the denominator was non-factorable? How would we approach the limit?
2. How do you evaluate limits with indeterminate forms like $\frac{0}{0}$?
3. What is the general process for finding limits at points of discontinuity?
4. How would the limit change if the numerator was a higher degree polynomial?
5. Can this method apply to limits approaching infinity?

Tip: Always try factoring the denominator first when dealing with rational expressions in limits. It often simplifies the calculation!