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In the image, you are asked to determine the set \( E \) and then to find the supremum (sup) and infimum (inf) of \( E \). I'll guide you through the highlighted cases.

### Highlighted Case 2:
\[ E = \left\{ \frac{(-1)^n + 1}{p} \mid (n,p) \in \mathbb{N}^* \times \mathbb{N}^* \right\} \]

- **Interpretation**:
  - \( n \) and \( p \) are natural numbers, excluding zero (i.e., \( \mathbb{N}^* = \mathbb{N} \setminus \{0\} \)).
  - \( (-1)^n \) alternates between \( 1 \) when \( n \) is even and \( -1 \) when \( n \) is odd.

Thus, the expression for elements of \( E \) simplifies as follows:
- When \( n \) is even (\( (-1)^n = 1 \)), the element becomes:
  \[
  \frac{1 + 1}{p} = \frac{2}{p}
  \]
- When \( n \) is odd (\( (-1)^n = -1 \)), the element becomes:
  \[
  \frac{-1 + 1}{p} = \frac{0}{p} = 0
  \]

Thus, the set \( E \) consists of \( 0 \) and \( \frac{2}{p} \) for every \( p \in \mathbb{N}^* \). So, 
\[
E = \left\{ 0, \frac{2}{1}, \frac{2}{2}, \frac{2}{3}, \dots \right\} = \left\{ 0 \right\} \cup \left\{ \frac{2}{p} \mid p \in \mathbb{N}^* \right\}
\]

- **Finding sup and inf**:
  - The **infimum** (inf) is the smallest value in the set, which is clearly \( 0 \).
  - The **supremum** (sup) is the largest value, which is the limit of \( \frac{2}{p} \) as \( p \to 1 \). Therefore, \( \sup E = 2 \).

### Highlighted Case 3:
\[ E = \left\{ \frac{x + 1}{x + 2} \mid x \in \mathbb{R} \text{ and } x \leq -3 \right\} \]

- **Interpretation**:
  - The expression \( \frac{x+1}{x+2} \) is defined for all real numbers \( x \leq -3 \).

Now, we analyze this function. Consider the behavior of \( \frac{x+1}{x+2} \):
- As \( x \to -\infty \), the fraction tends to \( 1 \), but since \( x \leq -3 \), the values remain less than \( 1 \).
- At \( x = -3 \):
  \[
  \frac{-3 + 1}{-3 + 2} = \frac{-2}{-1} = 2
  \]
  
Thus, the function is decreasing for \( x \leq -3 \), and its maximum occurs at \( x = -3 \) with the value \( 2 \), and the minimum value approaches \( 1 \) as \( x \to -\infty \).

- **Finding sup and inf**:
  - The **supremum** is the largest value, which occurs at \( x = -3 \), so \( \sup E = 2 \).
  - The **infimum** is the smallest value, which is approached as \( x \to -\infty \), so \( \inf E = 1 \).

Would you like to dive deeper into any of these calculations or need further clarification? Here are five related questions you could explore:

1. What is the role of \( n \) and \( p \) in the generalization of Case 2's set \( E \)?
2. Can you explain how the sequence \( \frac{2}{p} \) behaves as \( p \) increases?
3. How would the set \( E \) change in Case 3 if \( x \) was restricted to different intervals?
4. What is the significance of limits in determining the supremum and infimum?
5. How does the behavior of the function \( \frac{x+1}{x+2} \) change for different values of \( x \)?

**Tip**: When finding the supremum and infimum, always consider the boundary behavior of the set or function, as well as how it behaves as it approaches those boundaries.

Determine the set E and find its supremum and infimum in the given cases.

Explore how to find the supremum and infimum of sets derived from functions like E = { (-1)^n + 1 / p } and E = { (x + 1) / (x + 2) | x ∈ R and x ≤ -3 }. This detailed solution includes step-by-step analysis of these cases, making it suitable for undergraduate students studying algebra or analysis.

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