The value of `lim_(xrarr1)(xtan{x})/(x-1)` is equal to (where `{x}` denotes the fractional part of x)

Options

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

Answer

To solve the limit \( \lim_{x \to 1} \frac{x \tan(\{x\})}{x - 1} \), where \(\{x\}\) denotes the fractional part of \(x\), we will analyze the behavior of the function as \(x\) approaches 1 from both the left and the right.

### Step-by-step Solution:

1. **Understanding the Fractional Part**:
The fractional part of \(x\), denoted as \(\{x\}\), is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
For \(x\) approaching 1:
- If \(x < 1\), then \(\{x\} = x\).
- If \(x > 1\), then \(\{x\} = x - 1\).

2. **Finding the Left-Hand Limit (LHL)**:
We calculate the limit as \(x\) approaches 1 from the left (\(x \to 1^-\)):
\[ \lim_{x \to 1^-} \frac{x \tan(\{x\})}{x - 1} = \lim_{x \to 1^-} \frac{x \tan(x)}{x - 1} \]
As \(x\) approaches 1, \(x \to 1\) and \(\tan(x) \to \tan(1)\). The denominator approaches \(0\) from the negative side, leading to:
\[ \text{LHL} = \frac{1 \cdot \tan(1)}{0^-} = -\infty \]

3. **Finding the Right-Hand Limit (RHL)**:
Now we calculate the limit as \(x\) approaches 1 from the right (\(x \to 1^+\)):
\[ \lim_{x \to 1^+} \frac{x \tan(\{x\})}{x - 1} = \lim_{x \to 1^+} \frac{x \tan(x - 1)}{x - 1} \]
Here, as \(x\) approaches 1 from the right, \(x \to 1\) and \(\tan(x - 1) \to \tan(0) = 0\). The limit becomes:
\[ \text{RHL} = \frac{1 \cdot 0}{0^+} = 0 \]

4. **Conclusion**:
Since the left-hand limit (LHL) is \(-\infty\) and the right-hand limit (RHL) is \(0\), we conclude that:
\[ \lim_{x \to 1} \frac{x \tan(\{x\})}{x - 1} \text{ does not exist.} \]

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