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The limit problem from your image is:

\[
\lim_{x \to 0} \frac{\ln(3 - 2e^x)}{x}
\]

### To solve this, we can follow these steps:

1. **Substitute \( x = 0 \) directly:**
   \[
   \frac{\ln(3 - 2e^0)}{0} = \frac{\ln(3 - 2 \cdot 1)}{0} = \frac{\ln(1)}{0} = \frac{0}{0}
   \]
   This results in an indeterminate form \( \frac{0}{0} \), so we can apply **L'Hôpital's Rule**.

2. **Differentiate the numerator and denominator:**
   - The derivative of the numerator \( \ln(3 - 2e^x) \) is:
     \[
     \frac{d}{dx} \left[ \ln(3 - 2e^x) \right] = \frac{-2e^x}{3 - 2e^x}
     \]
   - The derivative of the denominator \( x \) is:
     \[
     \frac{d}{dx} [x] = 1
     \]

3. **Apply L'Hôpital's Rule:**
   \[
   \lim_{x \to 0} \frac{\ln(3 - 2e^x)}{x} = \lim_{x \to 0} \frac{-2e^x}{3 - 2e^x}
   \]
   Now, substitute \( x = 0 \):
   \[
   \frac{-2e^0}{3 - 2e^0} = \frac{-2 \cdot 1}{3 - 2 \cdot 1} = \frac{-2}{1} = -2
   \]

### Final Answer:
\[
\lim_{x \to 0} \frac{\ln(3 - 2e^x)}{x} = -2
\]

Would you like further details or explanations? 

Here are 5 related questions for deeper exploration:
1. What is L'Hôpital's Rule and when is it applied?
2. How can we differentiate logarithmic functions more generally?
3. What are other methods for solving limits that result in indeterminate forms?
4. How can the behavior of \( e^x \) as \( x \to 0 \) affect limit evaluations?
5. Can we use series expansions to solve limits like this one?

**Tip:** When facing indeterminate forms like \( \frac{0}{0} \), always check if L'Hôpital's Rule can be applied by differentiating the numerator and denominator.

Discover how to solve the limit lim(x→0) (ln(3 - 2e^x)) / x using L'Hopital's Rule and logarithmic differentiation. This problem focuses on handling indeterminate forms like 0/0 and explores key calculus concepts. Ideal for university-level calculus students.

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