Math Problem Statement

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

Solution

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:

    • en!ne \sqrt[n]{n!}
    • log(1+28n)\log \left( 1 + \frac{28}{n} \right)
    • (e1n12πn)1n\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:

    • limnnn!n=e\lim_{n \to \infty} n \sqrt[n]{n!} = e
    • limx0log(1+x)=1\lim_{x \to 0} \log(1 + x) = 1 when x0x \to 0, which can be useful here as 28n0\frac{28}{n} \to 0 when nn \to \infty.
    • limx0ex1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1, useful for simplifying the second part within the logarithmic term.
  3. Simplify log(1+28n)\log \left( 1 + \frac{28}{n} \right): For nn \to \infty, 28n0\frac{28}{n} \to 0. Using the approximation log(1+x)x\log(1 + x) \approx x when xx is close to 0, we get: log(1+28n)28n\log \left( 1 + \frac{28}{n} \right) \approx \frac{28}{n}

  4. Evaluate (e1n12πn)1n\left( \frac{e^{\frac{1}{n}} - 1}{\sqrt{2 \pi n}} \right)^{\frac{1}{n}}: Using the approximation e1n11ne^{\frac{1}{n}} - 1 \approx \frac{1}{n} as nn \to \infty, we get: e1n12πn1n2πn\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 1n\frac{1}{n}: (1n2πn)1n1as 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: 28n1\frac{28}{n} - 1 As nn \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 ]


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 ex1xe^x - 1 \approx x for small xx affect the accuracy of calculations?

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

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Math Problem Analysis

Mathematical Concepts

Limits
Factorials
Logarithmic functions
Exponential functions
Asymptotic behavior

Formulas

\( \lim_{x \to 0} \log(1 + x) = 1 \)
\( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \)
\( \lim_{n \to \infty} e \sqrt[n]{n!} = e \)

Theorems

Limits and asymptotic approximations

Suitable Grade Level

Grades 11-12