`lim_(x->1) (x-1){x}.` where {.) denotes the fractional part, is equal to:

Options

A. 1
B. 0
C. Does not exist
D. None of these

Answer

To solve the limit \( \lim_{x \to 1} (x - 1) \{x\} \), where \(\{x\}\) denotes the fractional part of \(x\), we can follow these steps:

### Step 1: Rewrite the Fractional Part
The fractional part of \(x\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \(x\).

### Step 2: Substitute the Definition into the Limit
Substituting the definition of the fractional part into the limit gives:
\[ \lim_{x \to 1} (x - 1)(x - \lfloor x \rfloor) \]

### Step 3: Analyze the Limit from the Left
To find the left-hand limit as \(x\) approaches 1 from the left (\(x \to 1^-\)):
- When \(x\) is slightly less than 1 (i.e., \(x = 1 - h\) where \(h \to 0^+\)), we have:
\[ \lfloor x \rfloor = 0 \]
Thus,
\[ \{x\} = (1 - h) - 0 = 1 - h \]
Now substituting into the limit:
\[ \lim_{h \to 0} (1 - h - 1)(1 - h) = \lim_{h \to 0} (-h)(1 - h) \]
As \(h \to 0\), this limit becomes:
\[ \lim_{h \to 0} -h(1 - h) = 0 \]

### Step 4: Analyze the Limit from the Right
Now, we find the right-hand limit as \(x\) approaches 1 from the right (\(x \to 1^+\)):
- When \(x\) is slightly more than 1 (i.e., \(x = 1 + h\) where \(h \to 0^+\)), we have:
\[ \lfloor x \rfloor = 1 \]
Thus,
\[ \{x\} = (1 + h) - 1 = h \]
Now substituting into the limit:
\[ \lim_{h \to 0} (1 + h - 1)(h) = \lim_{h \to 0} (h)(h) = \lim_{h \to 0} h^2 \]
As \(h \to 0\), this limit becomes:
\[ \lim_{h \to 0} h^2 = 0 \]

### Step 5: Conclude the Limits
Since both the left-hand limit and the right-hand limit equal 0, we conclude:
\[ \lim_{x \to 1} (x - 1) \{x\} = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]