he sequence of real number $ (x_n)$ is defined by $ x_1 \equal{} 0,$ $ x_2 \equal{} 2$ and $ x_{n\plus{}2} \equal{} 2^{\minus{}x_n} \plus{} \frac{1}{2}$ $ \forall n \equal{} 1,2,3 \ldots$ Prove that the sequence has a limit as $ n$ approaches $ \plus{}\infty.$ Determine the limit.
1. **Define the sequence and initial conditions:**
   The sequence \((x_n)\) is defined by:
   \[
   x_1 = 0, \quad x_2 = 2, \quad \text{and} \quad x_{n+2} = 2^{-x_n} + \frac{1}{2} \quad \forall n = 1, 2, 3, \ldots
   \]

2. **Assume the sequence has a limit \(L\):**
   Suppose the sequence \((x_n)\) converges to a limit \(L\) as \(n \to \infty\). Then, we have:
   \[
   \lim_{n \to \infty} x_n = L
   \]

3. **Use the recurrence relation at the limit:**
   Since \(x_{n+2} = 2^{-x_n} + \frac{1}{2}\), taking the limit on both sides as \(n \to \infty\), we get:
   \[
   L = 2^{-L} + \frac{1}{2}
   \]

4. **Solve the equation for \(L\):**
   To find \(L\), solve the equation:
   \[
   L = 2^{-L} + \frac{1}{2}
   \]
   Rearrange the equation:
   \[
   2L = 1 + 2^{-L}
   \]
   \[
   2L - 1 = 2^{-L}
   \]

5. **Check possible values for \(L\):**
   Let's test \(L = 1\):
   \[
   2(1) - 1 = 2^{-1}
   \]
   \[
   2 - 1 = \frac{1}{2}
   \]
   \[
   1 = \frac{1}{2}
   \]
   This is true, so \(L = 1\) is a solution.

6. **Verify the uniqueness of the solution:**
   To ensure that \(L = 1\) is the only solution, consider the function \(f(L) = 2L - 1 - 2^{-L}\). We need to show that \(f(L) = 0\) has a unique solution.
   \[
   f(L) = 2L - 1 - 2^{-L}
   \]
   Compute the derivative \(f'(L)\):
   \[
   f'(L) = 2 + 2^{-L} \ln(2)
   \]
   Since \(2 + 2^{-L} \ln(2) > 0\) for all \(L\), \(f(L)\) is strictly increasing. Therefore, \(f(L) = 0\) has at most one solution.

7. **Conclusion:**
   Since \(L = 1\) is a solution and \(f(L)\) is strictly increasing, \(L = 1\) is the unique solution.

\(\blacksquare\)

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