`lim_(x rarr0)(x[x])/(sin|x|)`, where `[.]` denote greater integer function

Options

A. is equal to 0
B. is equal to infinity
C. is equal to 1
D. does not exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{x \lfloor x \rfloor}{\sin |x|} \), where \( \lfloor . \rfloor \) denotes the greatest integer function, we will evaluate the left-hand limit and the right-hand limit separately.

### Step-by-Step Solution:

1. **Understanding the Greatest Integer Function**:
The greatest integer function \( \lfloor x \rfloor \) gives the largest integer less than or equal to \( x \).

2. **Finding the Left-Hand Limit**:
We need to evaluate \( \lim_{x \to 0^-} \frac{x \lfloor x \rfloor}{\sin |x|} \).
- For \( x < 0 \), \( \lfloor x \rfloor \) will be \( -1 \) when \( x \) is in the interval \( (-1, 0) \).
- Therefore, \( \lfloor x \rfloor = -1 \).
- The expression becomes:
\[ \lim_{x \to 0^-} \frac{x \cdot (-1)}{\sin(-x)} = \lim_{x \to 0^-} \frac{-x}{-\sin x} = \lim_{x \to 0^-} \frac{x}{\sin x} \]
- As \( x \to 0 \), we know that \( \frac{x}{\sin x} \to 1 \).
- Thus, the left-hand limit is:
\[ \lim_{x \to 0^-} \frac{x \lfloor x \rfloor}{\sin |x|} = 1 \]

3. **Finding the Right-Hand Limit**:
Now, we evaluate \( \lim_{x \to 0^+} \frac{x \lfloor x \rfloor}{\sin |x|} \).
- For \( x \geq 0 \), \( \lfloor x \rfloor = 0 \) when \( 0 \leq x < 1 \).
- Therefore, the expression becomes:
\[ \lim_{x \to 0^+} \frac{x \cdot 0}{\sin x} = \lim_{x \to 0^+} \frac{0}{\sin x} = 0 \]
- Thus, the right-hand limit is:
\[ \lim_{x \to 0^+} \frac{x \lfloor x \rfloor}{\sin |x|} = 0 \]

4. **Conclusion**:
Since the left-hand limit (1) is not equal to the right-hand limit (0), we conclude that the limit does not exist:
\[ \lim_{x \to 0} \frac{x \lfloor x \rfloor}{\sin |x|} \text{ does not exist.} \]

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