The given limit is:

\[
\lim_{x \to 3^+} \ln(x^2 - 9)
\]

### Step 1: Analyze the expression inside the logarithm
We have \( x^2 - 9 \), which can be factored as:

\[
x^2 - 9 = (x - 3)(x + 3)
\]

So, the expression becomes:

\[
\ln((x - 3)(x + 3))
\]

### Step 2: Examine the behavior as \( x \to 3^+ \)
As \( x \) approaches 3 from the right (\( x \to 3^+ \)), the term \( (x - 3) \) approaches 0 from the positive side, and \( (x + 3) \) approaches 6. Therefore, near \( x = 3 \), the product \( (x - 3)(x + 3) \) becomes small but positive.

Thus, as \( x \to 3^+ \), \( (x - 3)(x + 3) \to 0^+ \), and the logarithm of a value approaching 0 from the positive side approaches \( -\infty \).

### Conclusion:
\[
\lim_{x \to 3^+} \ln(x^2 - 9) = -\infty
\]

Let me know if you want further details or have any questions!

Here are 5 related questions for you to explore:
1. What happens to \( \lim_{x \to 3^-} \ln(x^2 - 9) \)?
2. How does the behavior change if the limit approaches from a negative direction?
3. What is the domain of the function \( \ln(x^2 - 9) \)?
4. How would you approach a limit problem involving \( \ln(f(x)) \) for more complex functions?
5. What would the limit be if we used \( \lim_{x \to 0^+} \ln(x^2 - 9) \)?

**Tip:** Always check the behavior of the function at critical points, especially for logarithmic expressions near 0, as they tend to approach \( -\infty \).

Explore the step-by-step solution to find the limit of ln(x^2 - 9) as x approaches 3 from the right. This problem covers factorization, logarithmic behavior, and limits, making it suitable for advanced high school students (Grades 11-12).

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