`lim_(xto1) (xsin(x-[x]))/(x-1)`, where `[.]` denotes the greatest integer function, is equal to

Options

A. `(2)/(pi-1)`
B. `(pi+1)/(2)`
C. `(2)/(pi+1)`
D. `(2(pi+1))/(pi-1)`

Explanation

To solve the limit \( \lim_{x \to 1} \frac{x \sin(x - [x])}{x - 1} \), where \([x]\) denotes the greatest integer function, we will analyze the behavior of the function as \(x\) approaches 1 from the left and right.

### Step-by-Step Solution:

1. **Understanding the Greatest Integer Function**:
- As \(x\) approaches 1 from the left (denoted as \(x \to 1^-\)), the greatest integer function \([x]\) will equal 0 because for any \(x < 1\), the greatest integer less than \(x\) is 0.

2. **Substituting into the Limit**:
- Therefore, for \(x\) approaching 1 from the left:
\[ x - [x] = x - 0 = x \]
- The limit can be rewritten as:
\[ \lim_{x \to 1^-} \frac{x \sin(x)}{x - 1} \]

3. **Evaluating the Numerator**:
- As \(x\) approaches 1, \(\sin(x)\) approaches \(\sin(1)\). Thus, the numerator \(x \sin(x)\) approaches \(1 \cdot \sin(1) = \sin(1)\).

4. **Evaluating the Denominator**:
- The denominator \(x - 1\) approaches \(0\) from the negative side (since \(x < 1\)), which means it approaches \(0^-\).

5. **Combining the Results**:
- Now we have:
\[ \lim_{x \to 1^-} \frac{\sin(1)}{x - 1} \]
- Since the numerator is a finite positive number (\(\sin(1)\)) and the denominator approaches \(0^-\), the limit approaches:
\[ -\infty \]

6. **Conclusion**:
- Since the left-hand limit approaches \(-\infty\), the overall limit does not exist.

### Final Answer:
\[ \lim_{x \to 1} \frac{x \sin(x - [x])}{x - 1} \text{ does not exist.} \]