If `a_(1)` is the greatest value of `f(x)` , where `f(x) =((1)/(2+[sinx]))` (where [.] denotes greatest integer function) and `a_(n-1)=((-1)^(n-2))/((n+1))+a_(n)` (where n in N), then `lim_(nrarroo) (a_(n))` is

Options

A. `1`
B. `e^(2)`
C. `In^(2)`
D. None of these

Explanation

To solve the problem, we need to analyze the given function and the recursive relation step by step.

### Step 1: Analyze the function \( f(x) = \frac{1}{2 + \lfloor \sin x \rfloor} \)

The greatest integer function \( \lfloor \sin x \rfloor \) takes the values based on the output of \( \sin x \). The sine function oscillates between -1 and 1. Therefore, the possible values for \( \lfloor \sin x \rfloor \) are:
- \( \lfloor \sin x \rfloor = -1 \) when \( \sin x \) is in the interval \([-1, 0)\)
- \( \lfloor \sin x \rfloor = 0 \) when \( \sin x = 0\)
- \( \lfloor \sin x \rfloor = 1 \) when \( \sin x \) is in the interval \([0, 1]\)

### Step 2: Determine the possible values of \( f(x) \)

Now, substituting these values into \( f(x) \):

1. When \( \lfloor \sin x \rfloor = -1 \):
\[ f(x) = \frac{1}{2 + (-1)} = \frac{1}{1} = 1 \]

2. When \( \lfloor \sin x \rfloor = 0 \):
\[ f(x) = \frac{1}{2 + 0} = \frac{1}{2} \]

3. When \( \lfloor \sin x \rfloor = 1 \):
\[ f(x) = \frac{1}{2 + 1} = \frac{1}{3} \]

### Step 3: Identify the greatest value of \( f(x) \)

From the calculations above, the maximum value of \( f(x) \) is:
\[ a_1 = 1 \]

### Step 4: Analyze the recursive relation \( a_{n-1} = \frac{(-1)^{n-2}}{(n+1)} + a_n \)

We need to find the limit of \( a_n \) as \( n \) approaches infinity.

### Step 5: Rearranging the recursive relation

Rearranging the recursive relation gives:
\[ a_n = a_{n-1} - \frac{(-1)^{n-2}}{(n+1)} \]

### Step 6: Calculate the limit as \( n \to \infty \)

As \( n \) approaches infinity, the term \( \frac{(-1)^{n-2}}{(n+1)} \) approaches 0. Therefore, the sequence \( a_n \) oscillates around a certain value.

To find the limit, we can observe that:
- For even \( n \), \( a_n \) will be slightly less than \( a_{n-1} \)
- For odd \( n \), \( a_n \) will be slightly more than \( a_{n-1} \)

Thus, the sequence converges to a limit \( L \) such that:
\[ L = L - 0 \]

This implies that the limit \( L \) is stable, and since the oscillation diminishes, we conclude:
\[ \lim_{n \to \infty} a_n = 1 \]

### Final Answer:
\[ \lim_{n \to \infty} a_n = 1 \]