The limit problem in the image is:

$\lim_{{x \to -3}} \frac{x^2 - 9}{x^2 - x - 12}$

Let's solve it step by step.

Step 1: Factor both the numerator and the denominator

• The numerator $x^2 - 9$ is a difference of squares: $x^2 - 9 = (x - 3)(x + 3)$

• The denominator $x^2 - x - 12$ can be factored by finding two numbers that multiply to $-12$ and add to $-1$. These numbers are $-4$ and $3$: $x^2 - x - 12 = (x - 4)(x + 3)$

Step 2: Simplify the expression

Now substitute the factored forms: $\frac{(x - 3)(x + 3)}{(x - 4)(x + 3)}$

Cancel out the common factor $(x + 3)$ from both the numerator and the denominator (as long as $x \neq -3$, which is true as we're taking a limit): $\frac{x - 3}{x - 4}$

Step 3: Evaluate the limit

Now, evaluate the limit as $x \to -3$: $\lim_{{x \to -3}} \frac{x - 3}{x - 4} = \frac{-3 - 3}{-3 - 4} = \frac{-6}{-7} = \frac{6}{7}$

Thus, the value of the limit is:

$\boxed{\frac{6}{7}}$

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

Here are 5 related questions to expand on this topic:

1. How do you handle limits when the expression includes a square root?
2. What is the difference between solving limits algebraically and using L'Hopital's Rule?
3. How would the solution change if the limit approached a different number?
4. Can you explain how limits apply to continuity in calculus?
5. What happens if direct substitution in the limit leads to a zero in the denominator?

Tip: When factoring quadratics, always check for common factoring techniques like difference of squares or simple factorization rules.