https://file.aiquickdraw.com/mathbot/file/1729563869099y7mj38t2.jpg

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?

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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.

lim (x^2 - 9) / (x^2 - x - 12) as x approaches -3

Learn how to calculate the limit of (x^2 - 9) / (x^2 - x - 12) as x approaches -3. This solution involves factoring quadratics and simplifying rational expressions. Suitable for students in Grades 10-12.

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