A function `f(x)` is such that `f(x+1)=sqrt(f(x)+2)`, where `x in I^+ uu {0}` and `f(0)=0` then
The value of `lim_(x rarroo) f(x)` is

Options

A. `sqrt2`
B. 2
C. `2sqrt2`
D. 4

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined by the recurrence relation \( f(x+1) = \sqrt{f(x) + 2} \) with the initial condition \( f(0) = 0 \).

### Step-by-Step Solution:

1. **Initial Condition**:
We start with the given condition:
\[ f(0) = 0 \]

2. **Finding \( f(1) \)**:
Using the recurrence relation for \( x = 0 \):
\[ f(1) = \sqrt{f(0) + 2} = \sqrt{0 + 2} = \sqrt{2} \]

3. **Finding \( f(2) \)**:
Now, using the recurrence relation for \( x = 1 \):
\[ f(2) = \sqrt{f(1) + 2} = \sqrt{\sqrt{2} + 2} \]

4. **Finding \( f(3) \)**:
Next, we calculate \( f(3) \) using \( x = 2 \):
\[ f(3) = \sqrt{f(2) + 2} = \sqrt{\sqrt{\sqrt{2} + 2} + 2} \]

5. **General Behavior**:
As we continue this process, we notice that each subsequent value of \( f(n) \) is defined in terms of the previous value plus 2, under a square root. To find the limit as \( x \) approaches infinity, we can denote:
\[ L = \lim_{x \to \infty} f(x) \]
Assuming this limit exists, we can substitute \( L \) into the recurrence relation:
\[ L = \sqrt{L + 2} \]

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

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

8. **Factoring the Quadratic**:
We can factor this equation:
\[ (L - 2)(L + 1) = 0 \]
This gives us two potential solutions:
\[ L = 2 \quad \text{or} \quad L = -1 \]

9. **Choosing the Valid Solution**:
Since \( f(x) \) is defined as a square root function, it must be non-negative. Therefore, we discard \( L = -1 \) and accept:
\[ L = 2 \]

10. **Conclusion**:
Thus, the value of \( \lim_{x \to \infty} f(x) \) is:
\[ \boxed{2} \]