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: $\sup A = 1$.

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.