Answer :

To evaluate the limit [tex]\(\lim_{x \rightarrow -3} \frac{x^2 - 9}{x^2 - x - 12}\)[/tex], let's follow these steps:

1. Factorize the numerator and the denominator:
- The numerator [tex]\(x^2 - 9\)[/tex] can be written as a difference of squares: [tex]\(x^2 - 9 = (x - 3)(x + 3)\)[/tex].
- The denominator [tex]\(x^2 - x - 12\)[/tex] can be factorized by finding two numbers that multiply to [tex]\(-12\)[/tex] and add to [tex]\(-1\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(3\)[/tex]. Therefore, the denominator can be factorized as: [tex]\(x^2 - x - 12 = (x - 4)(x + 3)\)[/tex].

2. Rewrite the fraction with the factorized expressions:
[tex]\[ \frac{x^2 - 9}{x^2 - x - 12} = \frac{(x - 3)(x + 3)}{(x - 4)(x + 3)} \][/tex]

3. Simplify the expression by canceling out common factors:
Since [tex]\((x + 3)\)[/tex] appears both in the numerator and the denominator, we can cancel it out (provided [tex]\(x \neq -3\)[/tex]):
[tex]\[ \frac{(x - 3)(x + 3)}{(x - 4)(x + 3)} = \frac{x - 3}{x - 4} \][/tex]

4. Evaluate the limit of the simplified expression as [tex]\(x\)[/tex] approaches [tex]\(-3\)[/tex]:
We now have:
[tex]\[ \lim_{x \rightarrow -3} \frac{x - 3}{x - 4} \][/tex]
Substitute [tex]\(x = -3\)[/tex] into the simplified expression:
[tex]\[ \frac{-3 - 3}{-3 - 4} = \frac{-6}{-7} = \frac{6}{7} \][/tex]

Thus, the limit is:
[tex]\[ \lim_{x \rightarrow -3} \frac{x^2 - 9}{x^2 - x - 12} = \frac{6}{7} \][/tex]