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The limit you're tasked with solving is:

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

Let's break it down:

### Step 1: Analyze the limit expression
The function inside the limit is \( \ln(9 - x) \). As \( x \to 9^- \) (i.e., approaching 9 from the left-hand side), the term \( 9 - x \) approaches 0 from the positive side because \( 9 - x > 0 \) for \( x < 9 \).

### Step 2: Behavior of the natural logarithm near zero
We know that:

\[
\lim_{y \to 0^+} \ln(y) = -\infty
\]

Since \( 9 - x \to 0^+ \) as \( x \to 9^- \), the expression \( \ln(9 - x) \) tends towards \( -\infty \).

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

Thus, the answer is:

\[
- \infty
\]

Would you like further details or explanations? Here are some related questions to expand your understanding:

1. What happens to \( \ln(x) \) as \( x \to \infty \)?
2. How would this limit change if \( x \to 9^+ \) instead of \( x \to 9^- \)?
3. What is the derivative of \( \ln(9 - x) \), and how does it behave as \( x \to 9 \)?
4. Can this limit be evaluated using L'Hôpital's Rule?
5. How does this behavior compare to the behavior of other logarithmic functions near their vertical asymptotes?

**Tip:** When analyzing limits involving logarithmic functions, always check the behavior near points where the argument tends to zero or infinity.

Find the limit of ln(9 - x) as x approaches 9 from the left-hand side (x → 9⁻).

Discover how to find the limit of ln(9 - x) as x approaches 9 from the left-hand side. This problem explores logarithmic functions and their behavior near vertical asymptotes, making it suitable for college-level calculus students.

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