The value of `lim_(xtoa)[sqrt(2-x)+sqrt(1+x)]`, where `a in[0,1/2]` and `[.]` denotes the greatest integer function is:

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the limit \( \lim_{x \to a} \left( \sqrt{2 - x} + \sqrt{1 + x} \right) \) where \( a \in [0, \frac{1}{2}] \) and using the greatest integer function, we can follow these steps:

### Step 1: Substitute the limit into the function
We start by substituting \( x = a \) into the expression:
\[ \sqrt{2 - a} + \sqrt{1 + a} \]

### Step 2: Analyze the range of \( a \)
Since \( a \) can take values in the interval \([0, \frac{1}{2}]\), we will evaluate the expression at the endpoints of this interval.

### Step 3: Evaluate at the endpoints
1. **When \( a = 0 \)**:
\[ \sqrt{2 - 0} + \sqrt{1 + 0} = \sqrt{2} + \sqrt{1} = \sqrt{2} + 1 \approx 2.414 \]

2. **When \( a = \frac{1}{2} \)**:
\[ \sqrt{2 - \frac{1}{2}} + \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{3}{2}} = 2\sqrt{\frac{3}{2}} \approx 2.449 \]

### Step 4: Determine the range of values
Now we have:
- At \( a = 0 \), the value is approximately \( 2.414 \).
- At \( a = \frac{1}{2} \), the value is approximately \( 2.449 \).

Thus, as \( a \) varies from \( 0 \) to \( \frac{1}{2} \), the expression \( \sqrt{2 - a} + \sqrt{1 + a} \) takes values in the interval \( [2.414, 2.449] \).

### Step 5: Apply the greatest integer function
The greatest integer function \( \lfloor x \rfloor \) gives the largest integer less than or equal to \( x \).

Since:
\[ 2.414 < 2.449 < 3 \]
The greatest integer function will yield:
\[ \lfloor \sqrt{2 - a} + \sqrt{1 + a} \rfloor = 2 \]
for all \( a \in [0, \frac{1}{2}] \).

### Conclusion
Thus, the value of the limit is:
\[ \lim_{x \to a} \left( \sqrt{2 - x} + \sqrt{1 + x} \right) = 2 \]

### Final Answer
\[ \boxed{2} \]