`lim_(x-(pi)/(2)) [([sinx]-[cosx]+1)/(3)]=` (where `[.]` denotes the greatest integer integer function)

Options

A. 0
B. 1
C. `-1`
D. does not exist

Explanation

To solve the limit \( \lim_{x \to \frac{\pi}{2}} \frac{[\sin x] - [\cos x] + 1}{3} \), where \([.]\) denotes the greatest integer function, we will analyze the left-hand limit (LHL) and the right-hand limit (RHL) separately.

### Step 1: Evaluate the Left-Hand Limit (LHL)

As \( x \) approaches \( \frac{\pi}{2} \) from the left (\( x < \frac{\pi}{2} \)):
- The value of \( \sin x \) approaches \( 1 \) (but is less than \( 1 \)).
- The value of \( \cos x \) approaches \( 0 \) (but is greater than \( -1 \)).

Thus:
- \( [\sin x] = 0 \) (since \( \sin x \) is in the range \( [0, 1) \)).
- \( [\cos x] = 0 \) (since \( \cos x \) is in the range \( [0, 1) \)).

Now substituting these values into the limit:
\[ \text{LHL} = \frac{0 - 0 + 1}{3} = \frac{1}{3} \]
Now, applying the greatest integer function:
\[ [\frac{1}{3}] = 0 \]

### Step 2: Evaluate the Right-Hand Limit (RHL)

As \( x \) approaches \( \frac{\pi}{2} \) from the right (\( x > \frac{\pi}{2} \)):
- The value of \( \sin x \) approaches \( 1 \) (but is still less than \( 1 \)).
- The value of \( \cos x \) approaches \( 0 \) (but is less than \( 0 \)).

Thus:
- \( [\sin x] = 0 \) (since \( \sin x \) is still in the range \( [0, 1) \)).
- \( [\cos x] = -1 \) (since \( \cos x \) is in the range \( (-1, 0) \)).

Now substituting these values into the limit:
\[ \text{RHL} = \frac{0 - (-1) + 1}{3} = \frac{0 + 1 + 1}{3} = \frac{2}{3} \]
Now, applying the greatest integer function:
\[ [\frac{2}{3}] = 0 \]

### Step 3: Conclusion

Since both the left-hand limit and the right-hand limit yield the same result:
\[ \text{LHL} = 0 \quad \text{and} \quad \text{RHL} = 0 \]
Thus, the limit exists and is equal to:
\[ \lim_{x \to \frac{\pi}{2}} \frac{[\sin x] - [\cos x] + 1}{3} = 0 \]

### Final Answer
\[ \boxed{0} \]