The problem asks to evaluate the following limit:

$\lim_{x \to -3} \frac{x^2 + x - 6}{x^2 - 9}.$

Step 1: Factorize the numerator and denominator

1. Numerator ($x^2 + x - 6$): Factorize: $x^2 + x - 6 = (x + 3)(x - 2).$

2. Denominator ($x^2 - 9$): Recognize as a difference of squares: $x^2 - 9 = (x + 3)(x - 3).$

The expression becomes: $\frac{x^2 + x - 6}{x^2 - 9} = \frac{(x + 3)(x - 2)}{(x + 3)(x - 3)}.$

Step 2: Simplify the expression

Cancel out the common term $(x + 3)$ from numerator and denominator, keeping in mind that $x \neq -3$ (to avoid division by zero): $\frac{(x + 3)(x - 2)}{(x + 3)(x - 3)} = \frac{x - 2}{x - 3}.$

Step 3: Substitute $x = -3$ into the simplified expression

Now evaluate the limit: $\lim_{x \to -3} \frac{x - 2}{x - 3} = \frac{-3 - 2}{-3 - 3} = \frac{-5}{-6} = \frac{5}{6}.$

Final Answer:

$\lim_{x \to -3} \frac{x^2 + x - 6}{x^2 - 9} = \frac{5}{6}.$

Let me know if you have any questions or want further clarifications!

5 Related Questions:

1. How can we handle limits involving indeterminate forms like $\frac{0}{0}$?
2. What is the difference of squares formula, and how is it applied here?
3. Can this limit be solved using L'Hôpital's Rule? If so, how?
4. How do we verify that the simplified form $\frac{x - 2}{x - 3}$ is valid for $x \neq -3$?
5. Why do we factorize the expression before evaluating the limit?

Tip: Always check for common factors in rational expressions when evaluating limits to simplify the computation.