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The problem asks to find the possible limit \( l \) for the recurrence sequence \( (u_n) \) defined as:

\[
u_{n+1} = f(u_n) \quad \text{with} \quad u_0 = 1 \quad \text{and} \quad f(x) = \sqrt{8 - 2x}.
\]

### Solution:

1. **Determine the limit condition**: If the sequence \( (u_n) \) converges to a limit \( l \), we must have:
   \[
   u_{n+1} = u_n = l.
   \]
   Therefore, the limit \( l \) must satisfy the equation:
   \[
   l = \sqrt{8 - 2l}.
   \]

2. **Solve for \( l \)**: Square both sides of the equation to eliminate the square root:
   \[
   l^2 = 8 - 2l.
   \]
   Rearrange the equation to form a standard quadratic equation:
   \[
   l^2 + 2l - 8 = 0.
   \]
   Solve this quadratic equation using the quadratic formula:
   \[
   l = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2}.
   \]
   Simplifying further:
   \[
   l = \frac{-2 \pm 6}{2}.
   \]
   This gives two possible solutions:
   \[
   l = \frac{-2 + 6}{2} = 2 \quad \text{or} \quad l = \frac{-2 - 6}{2} = -4.
   \]

3. **Analyze the solutions**: Since the function \( f(x) = \sqrt{8 - 2x} \) only produces non-negative values (because of the square root), the limit \( l = -4 \) is not valid. Thus, the only possible limit is:
   \[
   l = 2.
   \]

### Conclusion:
The only possible limit for the sequence \( (u_n) \) is \( l = 2 \).

---

Would you like further details or clarifications? Here are some related questions you could explore:

1. What happens to the sequence \( (u_n) \) for initial values other than \( u_0 = 1 \)?
2. How can we verify if \( u_n \) actually converges to the limit \( l = 2 \)?
3. Can this recurrence relation be solved explicitly, or is numerical computation required?
4. How would the behavior of the sequence change if the function \( f(x) \) were slightly modified?
5. What conditions on a recurrence sequence guarantee convergence to a limit?

**Tip**: Always check the domain of the function \( f(x) \) when solving recurrence relations to ensure valid solutions.

Soit la suite (u_n)_N définie par récurrence par u_{n+1} = f(u_n) avec u_0 = 1 et f(x) = \sqrt{8 - 2x}. Quelle est la seule limite possible l pour (u_n)_N?

This problem involves finding the possible limit of the recurrence sequence defined by u_{n+1} = sqrt(8 - 2u_n) with an initial condition of u_0 = 1. The solution involves solving a quadratic equation to determine the only valid limit.

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