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The question asks to determine, for each set \( A \), the minimum (\( \min A \)), infimum (\( \inf A \)), maximum (\( \max A \)), and supremum (\( \sup A \)) of \( A \), if they exist. Let's analyze each case step by step.

### 1. \( A = \left\{ \frac{1}{n} \mid n \in \mathbb{N}^* \right\} \)

#### Analysis:
- \( \frac{1}{n} \) is a strictly decreasing sequence as \( n \) increases.
- As \( n \to \infty \), \( \frac{1}{n} \to 0 \).
- The set contains positive terms, starting from \( \frac{1}{1} = 1 \) and getting smaller as \( n \) increases.

#### Results:
- **Minimum**: \( \min A = \frac{1}{1} = 1 \) (since \( \frac{1}{n} \) is defined for \( n \geq 1 \)).
- **Infimum**: \( \inf A = 0 \) (since the terms approach 0 but never reach it).
- **Maximum**: \( \max A = \min A = 1 \) (same as the minimum, because the sequence is decreasing).
- **Supremum**: \( \sup A = 1 \) (since the largest element in the set is 1).

### 2. \( A = \left\{ \frac{(-1)^n}{n} + \frac{2}{n} \mid n \in \mathbb{N} \right\} \)

#### Analysis:
- This expression can be written as:
  \[
  A = \left\{ \frac{(-1)^n + 2}{n} \mid n \in \mathbb{N} \right\}.
  \]
- For odd \( n \), \( (-1)^n = -1 \), so the expression becomes:
  \[
  A = \left\{ \frac{-1 + 2}{n} = \frac{1}{n} \mid n \text{ is odd} \right\}.
  \]
- For even \( n \), \( (-1)^n = 1 \), so the expression becomes:
  \[
  A = \left\{ \frac{1 + 2}{n} = \frac{3}{n} \mid n \text{ is even} \right\}.
  \]
- As \( n \to \infty \), both \( \frac{1}{n} \) and \( \frac{3}{n} \) tend to 0.

#### Results:
- **Minimum**: \( \min A = \frac{1}{1} = 1 \).
- **Infimum**: \( \inf A = 0 \) (since the terms approach 0 as \( n \to \infty \)).
- **Maximum**: \( \max A = \frac{3}{2} \) (since \( \frac{3}{2} \) occurs when \( n = 2 \), the largest term for even \( n \)).
- **Supremum**: \( \sup A = \frac{3}{2} \) (the largest possible value of the set).

### 3. \( A = \left\{ \frac{1 - \frac{1}{n}}{1 + \frac{1}{n}} \mid n \in \mathbb{N}^* \right\} \)

#### Simplifying the expression:
\[
A = \left\{ \frac{\frac{n - 1}{n}}{\frac{n + 1}{n}} = \frac{n - 1}{n + 1} \mid n \in \mathbb{N}^* \right\}.
\]
- As \( n \to \infty \), \( \frac{n - 1}{n + 1} \to 1 \).
- For small \( n \), calculate:
  - When \( n = 1 \), \( A = \frac{0}{2} = 0 \).
  - When \( n = 2 \), \( A = \frac{1}{3} \).
  - The sequence increases as \( n \) increases.

#### Results:
- **Minimum**: \( \min A = 0 \) (attained when \( n = 1 \)).
- **Infimum**: \( \inf A = 0 \).
- **Maximum**: The sequence does not reach 1 but approaches it as \( n \to \infty \), so there is no maximum.
- **Supremum**: \( \sup A = 1 \).

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Would you like further details or clarifications? Here are five questions you might explore next:

1. How is the supremum different from the maximum of a set?
2. What happens when a sequence alternates like in case 2?
3. Can you describe how limits relate to infimum and supremum?
4. How would the set in case 1 change if the domain was extended?
5. What are the characteristics of a set that makes its supremum and maximum equal?

**Tip**: Always verify the behavior of sequences as \( n \to \infty \) to determine the infimum and supremum effectively.

In each of the following cases, determine, if they exist, the minimum (min A), infimum (inf A), maximum (max A), and supremum (sup A) of A.

This problem explores the minimum, infimum, maximum, and supremum of three different mathematical sequences. Learn how to analyze sequences such as 1/n, ((-1)^n)/n + 2/n, and (1 - 1/n)/(1 + 1/n) to find these values using limits and set theory.

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