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The limit presented is:

\[
\lim_{{x \to -3}} \frac{{x^2 - 9}}{{x^2 + 9}}
\]

### Step-by-step solution:

1. **Simplify the expression**:
   \[
   \frac{{x^2 - 9}}{{x^2 + 9}} = \frac{{(x - 3)(x + 3)}}{{x^2 + 9}}
   \]
   The numerator is factored as a difference of squares.

2. **Substitute \(x = -3\)** into the simplified expression:
   \[
   \frac{{(-3 - 3)(-3 + 3)}}{{(-3)^2 + 9}} = \frac{{(-6)(0)}}{{9 + 9}} = 0
   \]

Since the numerator becomes 0 and the denominator remains non-zero, the value of the limit is:

\[
\boxed{0}
\]

Would you like further details on any of the steps, or do you have any specific questions?

Here are some related questions to explore:
1. How do you approach limits involving indeterminate forms?
2. What is the difference between one-sided and two-sided limits?
3. How can the difference of squares help in simplifying limits?
4. What if the denominator approached 0 as well? How would that change the process?
5. How can you apply L'Hopital's Rule to find limits?

**Tip:** Always check if you can simplify the expression before plugging in the value to avoid indeterminate forms!

This limit problem involves evaluating lim(x -> -3) of (x^2 - 9)/(x^2 + 9). By factoring the numerator as a difference of squares and substituting x = -3, the limit simplifies to 0. This is a key concept in calculus for understanding the behavior of functions as they approach specific points.

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