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The problem asks to compute the limit:

\[
\lim_{n \to \infty} \frac{b_n}{b_{n+1}},
\]

where \( b_n = (3n - 2)^n \).

### Step-by-step solution:

1. **Expression for \( b_{n+1} \):**
   Using the given expression \( b_n = (3n - 2)^n \), we compute \( b_{n+1} \):
   
   \[
   b_{n+1} = (3(n+1) - 2)^{n+1} = (3n + 3 - 2)^{n+1} = (3n + 1)^{n+1}.
   \]

2. **Set up the limit:**
   We now substitute \( b_n \) and \( b_{n+1} \) into the given limit expression:

   \[
   \lim_{n \to \infty} \frac{b_n}{b_{n+1}} = \lim_{n \to \infty} \frac{(3n - 2)^n}{(3n + 1)^{n+1}}.
   \]

3. **Simplify the expression:**
   Let's simplify the limit:

   \[
   \frac{b_n}{b_{n+1}} = \frac{(3n - 2)^n}{(3n + 1)^{n+1}} = \frac{(3n - 2)^n}{(3n + 1)^n \cdot (3n + 1)}.
   \]
   
   This can be rewritten as:

   \[
   \frac{b_n}{b_{n+1}} = \frac{(3n - 2)^n}{(3n + 1)^n} \cdot \frac{1}{3n + 1}.
   \]
   
   We now focus on the first fraction \( \frac{(3n - 2)^n}{(3n + 1)^n} \), which can be simplified as:

   \[
   \left( \frac{3n - 2}{3n + 1} \right)^n.
   \]

4. **Asymptotic behavior of the fraction:**
   For large \( n \), we can approximate:

   \[
   \frac{3n - 2}{3n + 1} \approx \frac{3n}{3n} = 1.
   \]

   Therefore,

   \[
   \left( \frac{3n - 2}{3n + 1} \right)^n \approx \left( 1 - \frac{3}{3n + 1} \right)^n.
   \]
   
   Using the approximation \( \left( 1 - \frac{a}{n} \right)^n \approx e^{-a} \) for large \( n \), we have:

   \[
   \left( 1 - \frac{3}{3n + 1} \right)^n \approx e^{-1}.
   \]

5. **Final limit:**
   Now, the overall limit becomes:

   \[
   \lim_{n \to \infty} \frac{b_n}{b_{n+1}} \approx \frac{e^{-1}}{3n + 1}.
   \]
   
   As \( n \to \infty \), the term \( \frac{1}{3n + 1} \) tends to 0, so the entire expression tends to 0. Therefore, the limit is:

   \[
   \lim_{n \to \infty} \frac{b_n}{b_{n+1}} = 0.
   \]

### Final Answer:
\[
\lim_{n \to \infty} \frac{b_n}{b_{n+1}} = 0.
\]

Would you like more details on any of the steps? Here are 5 related questions:

1. How does the approximation \( \left( 1 - \frac{a}{n} \right)^n \approx e^{-a} \) work?
2. Can you explain why \( \frac{3n - 2}{3n + 1} \) tends to 1 as \( n \to \infty \)?
3. What other methods can be used to solve limits involving sequences like this one?
4. What happens if the terms inside the sequence were different, such as \( (2n - 1)^n \)?
5. How do we deal with indeterminate forms in limits involving infinity?

### Tip:
When dealing with limits of sequences, recognizing standard forms and applying approximations like \( (1 - \frac{a}{n})^n \to e^{-a} \) can greatly simplify the process.

Calcular \( \lim_{n \to \infty} \frac{b_n}{b_{n+1}} \), donde \( b_n = (3n - 2)^n \).

This page explains how to calculate the limit \( \lim_{n \to \infty} \frac{b_n}{b_{n+1}} \), where \( b_n = (3n - 2)^n \). The problem involves applying asymptotic approximations for large n and using exponential approximations for sequences.

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