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To solve the given problem, we need to analyze the limit:

\[
\lim_{n \to \infty} e \sqrt[n]{n!} \left[ \log \left( 1 + \frac{28}{n} \right) - \left( \frac{e^{\frac{1}{n}} - 1}{\sqrt{2 \pi n}} \right)^{\frac{1}{n}} \right]
\]

### Step-by-Step Solution Outline:

1. **Examine Each Part of the Expression Separately**:
   We have three main components in the expression:
   - \( e \sqrt[n]{n!} \)
   - \( \log \left( 1 + \frac{28}{n} \right) \)
   - \( \left( \frac{e^{\frac{1}{n}} - 1}{\sqrt{2 \pi n}} \right)^{\frac{1}{n}} \)

2. **Apply Known Limits**:
   Based on the information given, let’s use the limits provided in the problem statement:
   - \(\lim_{n \to \infty} n \sqrt[n]{n!} = e\)
   - \(\lim_{x \to 0} \log(1 + x) = 1\) when \(x \to 0\), which can be useful here as \(\frac{28}{n} \to 0\) when \(n \to \infty\).
   - \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\), useful for simplifying the second part within the logarithmic term.

3. **Simplify \(\log \left( 1 + \frac{28}{n} \right)\)**:
   For \( n \to \infty \), \(\frac{28}{n} \to 0\). Using the approximation \(\log(1 + x) \approx x\) when \(x\) is close to 0, we get:
   \[
   \log \left( 1 + \frac{28}{n} \right) \approx \frac{28}{n}
   \]

4. **Evaluate \(\left( \frac{e^{\frac{1}{n}} - 1}{\sqrt{2 \pi n}} \right)^{\frac{1}{n}}\)**:
   Using the approximation \( e^{\frac{1}{n}} - 1 \approx \frac{1}{n} \) as \(n \to \infty\), we get:
   \[
   \frac{e^{\frac{1}{n}} - 1}{\sqrt{2 \pi n}} \approx \frac{1}{n \sqrt{2 \pi n}}
   \]
   Then, raising this expression to the power of \(\frac{1}{n}\):
   \[
   \left( \frac{1}{n \sqrt{2 \pi n}} \right)^{\frac{1}{n}} \to 1 \quad \text{as } n \to \infty
   \]

5. **Combine Terms and Take the Limit**:
   Substituting back, the expression inside the brackets becomes:
   \[
   \frac{28}{n} - 1
   \]
   As \(n \to \infty\), this entire expression approaches 0.

Thus, the final answer is:

\[
\lim_{n \to \infty} e \sqrt[n]{n!} \left[ \log \left( 1 + \frac{28}{n} \right) - \left( \frac{e^{\frac{1}{n}} - 1}{\sqrt{2 \pi n}} \right)^{\frac{1}{n}} \right] = 0
\]

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Would you like further details on any step, or do you have questions about this approach?

Here are some related questions you might find interesting:
1. How does Stirling’s approximation relate to limits involving factorials?
2. What other approximations are commonly used for logarithmic functions near 0?
3. Can this approach be extended to more complex expressions involving nested limits?
4. What are the convergence properties of limits involving factorials?
5. How does the approximation \( e^x - 1 \approx x \) for small \( x \) affect the accuracy of calculations?

**Tip:** Always check the behavior of each component as \(n \to \infty\) when dealing with complex limits, breaking down the expression step by step.

Using the given information, find the value of \( \lim_{n \to \infty} e \sqrt[n]{n!} \left[ \log \left( 1 + \frac{28}{n} \right) - \left( \frac{e^{\frac{1}{n}} - 1}{\sqrt{2 \pi n}} \right)^{\frac{1}{n}} \right] \).

Explore a detailed solution for finding the limit of a complex expression involving factorials, logarithmic functions, and exponential approximations. This problem utilizes advanced limit properties, asymptotic behavior, and is suitable for students in Grades 11-12.

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