Given $u_0,u_1$ with $0<u_0,u_1<1$, define the sequence $(u_n)$ recurrently by the formula
$$u_{n+2}=\frac12\left(\sqrt{u_{n+1}}+\sqrt{u_n}\right).$$(a) Prove that the sequence $u_n$ is convergent and find its limit.
(b) Prove that, starting from some index $n_0$, the sequence $u_n$ is monotonous.
(a) To prove that the sequence \( u_n \) is convergent and find its limit, we will follow these steps:

1. **Boundedness and Monotonicity**:
   - First, we show that the sequence \( u_n \) is bounded above by 1. We use induction to prove this.
   - Base case: \( u_0 < 1 \) and \( u_1 < 1 \) (given).
   - Inductive step: Assume \( u_n < 1 \) and \( u_{n+1} < 1 \). We need to show \( u_{n+2} < 1 \).
     \[
     u_{n+2} = \frac{1}{2} \left( \sqrt{u_{n+1}} + \sqrt{u_n} \right)
     \]
     Since \( 0 < u_n < 1 \) and \( 0 < u_{n+1} < 1 \), we have \( 0 < \sqrt{u_n} < 1 \) and \( 0 < \sqrt{u_{n+1}} < 1 \). Therefore,
     \[
     \sqrt{u_n} + \sqrt{u_{n+1}} < 2
     \]
     Dividing by 2, we get
     \[
     \frac{1}{2} \left( \sqrt{u_{n+1}} + \sqrt{u_n} \right) < 1
     \]
     Hence, \( u_{n+2} < 1 \). By induction, \( u_n < 1 \) for all \( n \).

2. **Convergence**:
   - Since \( 0 < u_n < 1 \) for all \( n \), the sequence is bounded.
   - To show that the sequence is monotonic, we need to prove that it is either non-increasing or non-decreasing from some index \( n_0 \).
   - Consider the function \( f(x) = \frac{1}{2} \left( \sqrt{x} + \sqrt{u_{n-1}} \right) \). This function is continuous and strictly increasing for \( x \in (0, 1) \).
   - If \( u_n \leq u_{n+1} \), then \( u_{n+1} \leq u_{n+2} \) because
     \[
     u_{n+2} = \frac{1}{2} \left( \sqrt{u_{n+1}} + \sqrt{u_n} \right) \geq \frac{1}{2} \left( \sqrt{u_n} + \sqrt{u_n} \right) = u_n
     \]
   - Similarly, if \( u_n \geq u_{n+1} \), then \( u_{n+1} \geq u_{n+2} \).
   - Therefore, the sequence \( u_n \) is eventually monotonic.

3. **Finding the Limit**:
   - Let \( L \) be the limit of the sequence \( u_n \). Since the sequence is convergent, we have
     \[
     \lim_{n \to \infty} u_n = L
     \]
   - Taking the limit on both sides of the recurrence relation, we get
     \[
     L = \frac{1}{2} \left( \sqrt{L} + \sqrt{L} \right)
     \]
     Simplifying, we obtain
     \[
     L = \sqrt{L}
     \]
   - Solving \( L = \sqrt{L} \), we get \( L^2 = L \), which gives \( L(L - 1) = 0 \). Therefore, \( L = 0 \) or \( L = 1 \).
   - Since \( 0 < u_n < 1 \) for all \( n \), the limit \( L \) must be within this interval. Hence, \( L = 1 \).

The final answer is \( \boxed{1} \)