If `{x}` denotes the fractional part of x, then `lim_(xrarr1) (x sin {x})/(x-1)`, is

Options

A. 0
B. -1
C. non-extistent
D. none of these

Explanation

To solve the limit problem \( \lim_{x \to 1} \frac{x \sin \{x\}}{x - 1} \), where \(\{x\}\) denotes the fractional part of \(x\), we will analyze the limit from both the right-hand side and the left-hand side.

### Step-by-Step Solution:

1. **Right-Hand Limit**:
We will first calculate the right-hand limit as \(x\) approaches 1 from the right, i.e., \(x \to 1^+\).
- Let \(x = 1 + h\) where \(h \to 0^+\).
- The fractional part \(\{x\} = \{1 + h\} = h\).
- Substitute into the limit:
\[ \lim_{h \to 0^+} \frac{(1 + h) \sin(h)}{(1 + h) - 1} = \lim_{h \to 0^+} \frac{(1 + h) \sin(h)}{h} \]
- This can be rewritten as:
\[ \lim_{h \to 0^+} \left( (1 + h) \cdot \frac{\sin(h)}{h} \right) \]
- As \(h \to 0\), \(\frac{\sin(h)}{h} \to 1\), so:
\[ \lim_{h \to 0^+} (1 + h) \cdot 1 = 1 + 0 = 1 \]

2. **Left-Hand Limit**:
Now, we calculate the left-hand limit as \(x\) approaches 1 from the left, i.e., \(x \to 1^-\).
- Let \(x = 1 - h\) where \(h \to 0^+\).
- The fractional part \(\{x\} = \{1 - h\} = 1 - h\).
- Substitute into the limit:
\[ \lim_{h \to 0^+} \frac{(1 - h) \sin(1 - h)}{(1 - h) - 1} = \lim_{h \to 0^+} \frac{(1 - h) \sin(1 - h)}{-h} \]
- This can be rewritten as:
\[ \lim_{h \to 0^+} -\frac{(1 - h) \sin(1 - h)}{h} \]
- As \(h \to 0\), \(\sin(1 - h) \to \sin(1)\), so:
\[ \lim_{h \to 0^+} -\frac{(1 - h) \sin(1)}{h} = -\infty \]

3. **Conclusion**:
Since the right-hand limit is \(1\) and the left-hand limit is \(-\infty\), we conclude that the limits do not match:
\[ \lim_{x \to 1} \frac{x \sin \{x\}}{x - 1} \text{ does not exist.} \]

### Final Answer:
The limit does not exist.