`lim_(xrarrpi)([sinx]+x)` is

Options

A. `pi`
B. `1+pi`
C. `pi-1`
D. does not exist

Answer

To solve the limit \( \lim_{x \to \pi} (\lfloor \sin x \rfloor + x) \), we will analyze the behavior of the function as \( x \) approaches \( \pi \).

### Step-by-Step Solution:

1. **Identify the components of the limit**:
We need to evaluate \( \lfloor \sin x \rfloor \) and \( x \) as \( x \) approaches \( \pi \).

2. **Evaluate \( \sin x \) as \( x \to \pi \)**:
We know that \( \sin \pi = 0 \). Therefore, as \( x \) approaches \( \pi \), \( \sin x \) approaches \( 0 \).

3. **Determine \( \lfloor \sin x \rfloor \)**:
Since \( \sin x \) approaches \( 0 \), we need to consider the behavior of \( \sin x \) just before \( x \) reaches \( \pi \). For values of \( x \) slightly less than \( \pi \), \( \sin x \) will be slightly positive but very close to \( 0 \). Thus, \( \lfloor \sin x \rfloor \) will be \( -1 \) because the greatest integer less than or equal to a small positive number approaching \( 0 \) is \( -1 \).

4. **Evaluate \( x \) as \( x \to \pi \)**:
As \( x \) approaches \( \pi \), \( x \) itself approaches \( \pi \).

5. **Combine the results**:
Now we can combine the results:
\[ \lfloor \sin x \rfloor + x \to -1 + \pi \]

6. **Calculate the limit**:
Therefore, we have:
\[ \lim_{x \to \pi} (\lfloor \sin x \rfloor + x) = -1 + \pi \]

### Final Answer:
\[ \lim_{x \to \pi} (\lfloor \sin x \rfloor + x) = \pi - 1 \]