If `x_(1)=3` and `x_(n+1)=sqrt(2+x_(n))" ",nge1,` then `lim_(ntooo) x_(n)`is

Options

A. 0
B. `(1)/(2)`
C. log2
D. `e^(4)`

Explanation

To solve the problem step by step, we start with the given recurrence relation and initial condition:

1. **Given Information**:
- \( x_1 = 3 \)
- \( x_{n+1} = \sqrt{2 + x_n} \) for \( n \geq 1 \)

2. **Finding the Limit**:
- We want to find \( \lim_{n \to \infty} x_n \). Let's denote this limit as \( L \).
- As \( n \) approaches infinity, both \( x_n \) and \( x_{n+1} \) approach the same limit \( L \). Therefore, we can write:
\[ L = \sqrt{2 + L} \]

3. **Squaring Both Sides**:
- To eliminate the square root, we square both sides of the equation:
\[ L^2 = 2 + L \]

4. **Rearranging the Equation**:
- Rearranging the equation gives us a standard quadratic equation:
\[ L^2 - L - 2 = 0 \]

5. **Factoring the Quadratic**:
- We can factor this quadratic equation:
\[ (L - 2)(L + 1) = 0 \]

6. **Finding the Roots**:
- Setting each factor to zero gives us the possible solutions:
\[ L - 2 = 0 \quad \Rightarrow \quad L = 2 \]
\[ L + 1 = 0 \quad \Rightarrow \quad L = -1 \]

7. **Determining the Valid Limit**:
- Since \( L \) represents the limit of the sequence \( x_n \) and the terms of the sequence are generated from square roots (which are non-negative), we discard \( L = -1 \) as it is not a valid solution.
- Thus, we conclude:
\[ L = 2 \]

8. **Final Result**:
- Therefore, the limit is:
\[ \lim_{n \to \infty} x_n = 2 \]