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!