1. **Problem statement:** We have a sequence defined by the recurrence relation $$u_0 = 11$$ $$u_{n+1} = \sqrt{u_n - 2} + 2, \quad n \in \mathbb{N}$$ We want to analyze the behavior of this sequence and find its limit. 2. **Function definition:** The function associated with the recurrence is $$f(x) = \sqrt{x - 2} + 2, \quad x \in [2, +\infty[\,.$$ 3. **Goal:** Show that the sequence $(u_n)$ converges and find its limit. 4. **Step 1: Fixed point analysis** We look for a fixed point $L$ such that $$L = \sqrt{L - 2} + 2.$$ 5. **Step 2: Solve for $L$** Subtract 2 from both sides: $$L - 2 = \sqrt{L - 2}.$$ Let $y = L - 2$, then $$y = \sqrt{y}.$$ Square both sides: $$y^2 = y.$$ 6. **Step 3: Solve quadratic equation** $$y^2 - y = 0$$ $$y(y - 1) = 0.$$ So, $$y = 0 \quad \text{or} \quad y = 1.$$ 7. **Step 4: Find corresponding $L$ values** Since $y = L - 2$, then $$L = 2 + y.$$ So, $$L = 2 + 0 = 2 \quad \text{or} \quad L = 2 + 1 = 3.$$ 8. **Step 5: Check stability and initial value** Given $u_0 = 11$, which is greater than 3, and the recurrence relation, the sequence tends to approach the stable fixed point. The problem states and proves that $$\lim_{n \to +\infty} u_n = 3.$$ 9. **Step 6: Inequalities and convergence** The problem provides the inequality $$0 \leq u_n - 3 \leq 8 \left( \frac{1}{2} \right)^n,$$ which shows that the difference between $u_n$ and 3 decreases exponentially, confirming convergence to 3. **Final answer:** $$\boxed{\lim_{n \to +\infty} u_n = 3}.$$