The value of `lim_(xto0)((tan({x}-1))sin{x})/({x}({x}-1)` is where `{x}` denotes the fractional part function

Options

A. 1
B. `tan 1`
C. `sin 1`
D. Doesn't exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{\tan(\{x\} - 1) \sin(\{x\})}{\{x\}(\{x\} - 1)} \), where \( \{x\} \) denotes the fractional part of \( x \), we will analyze the limit from both sides: the left-hand limit and the right-hand limit.

### Step 1: Understanding the fractional part function
The fractional part function \( \{x\} = x - \lfloor x \rfloor \). As \( x \) approaches 0, we need to consider two cases:
- When \( x \to 0^- \) (approaching from the left)
- When \( x \to 0^+ \) (approaching from the right)

### Step 2: Left-hand limit \( \lim_{x \to 0^-} \)
When \( x \to 0^- \):
- \( \{x\} = x - (-1) = x + 1 \) (since \( \lfloor x \rfloor = -1 \))
- Thus, \( \{x\} \to 1 \)

Substituting into the limit:
\[ \lim_{x \to 0^-} \frac{\tan(\{x\} - 1) \sin(\{x\})}{\{x\}(\{x\} - 1)} = \lim_{x \to 0^-} \frac{\tan((x + 1) - 1) \sin(x + 1)}{(x + 1)((x + 1) - 1)} \]
This simplifies to:
\[ \lim_{x \to 0^-} \frac{\tan(x) \sin(x + 1)}{(x + 1)(x)} = \lim_{x \to 0^-} \frac{\tan(x) \sin(x + 1)}{x(x + 1)} \]
As \( x \to 0^- \), \( \tan(x) \to x \) and \( \sin(x + 1) \to \sin(1) \):
\[ = \lim_{x \to 0^-} \frac{x \sin(1)}{x(x + 1)} = \lim_{x \to 0^-} \frac{\sin(1)}{x + 1} = \frac{\sin(1)}{1} = \sin(1) \]

### Step 3: Right-hand limit \( \lim_{x \to 0^+} \)
When \( x \to 0^+ \):
- \( \{x\} = x \) (since \( \lfloor x \rfloor = 0 \))
- Thus, \( \{x\} \to 0 \)

Substituting into the limit:
\[ \lim_{x \to 0^+} \frac{\tan(\{x\} - 1) \sin(\{x\})}{\{x\}(\{x\} - 1)} = \lim_{x \to 0^+} \frac{\tan(x - 1) \sin(x)}{x(x - 1)} \]
As \( x \to 0^+ \), \( \tan(x - 1) \to \tan(-1) \) and \( \sin(x) \to 0 \):
\[ = \lim_{x \to 0^+} \frac{\tan(-1) \cdot 0}{x(x - 1)} = 0 \]

### Step 4: Conclusion
Since the left-hand limit \( \lim_{x \to 0^-} = \sin(1) \) and the right-hand limit \( \lim_{x \to 0^+} = 0 \) are not equal, we conclude that:
\[ \lim_{x \to 0} \frac{\tan(\{x\} - 1) \sin(\{x\})}{\{x\}(\{x\} - 1)} \text{ does not exist.} \]