`lim_(xto0) [(sin(sgn(x)))/((sgn(x)))],` where `[.]` denotes the greatest integer function, is equal to

Options

A. `^(2n)p_(n)`
B. `^(2n)C_(n)`
C. `(2n)!`
D. none of these

Explanation

To solve the limit \( \lim_{x \to 0} \left( \frac{\sin(\text{sgn}(x))}{\text{sgn}(x)} \right) \), where \(\text{sgn}(x)\) is the signum function and \([.]\) denotes the greatest integer function, we will analyze the behavior of the function as \(x\) approaches 0 from both the right and the left.

### Step 1: Right-hand limit as \(x \to 0^+\)
- When \(x\) approaches 0 from the right (\(x \to 0^+\)), \(\text{sgn}(x) = 1\).
- Therefore, we can substitute this into our limit:
\[ \lim_{x \to 0^+} \frac{\sin(\text{sgn}(x))}{\text{sgn}(x)} = \frac{\sin(1)}{1} \]
- The value of \(\sin(1)\) is approximately \(0.8415\) (in radians).

### Step 2: Greatest integer function
- Since \(\sin(1) \approx 0.8415\), we need to evaluate the greatest integer function:
\[ [\sin(1)] = [0.8415] = 0 \]

### Step 3: Left-hand limit as \(x \to 0^-\)
- When \(x\) approaches 0 from the left (\(x \to 0^-\)), \(\text{sgn}(x) = -1\).
- Thus, we can substitute this into our limit:
\[ \lim_{x \to 0^-} \frac{\sin(\text{sgn}(x))}{\text{sgn}(x)} = \frac{\sin(-1)}{-1} \]
- We know that \(\sin(-1) = -\sin(1)\), so:
\[ \frac{-\sin(1)}{-1} = \sin(1) \]
- Again, \(\sin(1) \approx 0.8415\).

### Step 4: Greatest integer function for left-hand limit
- Similar to the right-hand limit, we evaluate the greatest integer function:
\[ [\sin(1)] = [0.8415] = 0 \]

### Step 5: Conclusion
- Since both the right-hand limit and left-hand limit yield the same result:
\[ \lim_{x \to 0^+} \left( \frac{\sin(\text{sgn}(x))}{\text{sgn}(x)} \right) = 0 \]
\[ \lim_{x \to 0^-} \left( \frac{\sin(\text{sgn}(x))}{\text{sgn}(x)} \right) = 0 \]
- Therefore, the overall limit is:
\[ \lim_{x \to 0} \left( \frac{\sin(\text{sgn}(x))}{\text{sgn}(x)} \right) = 0 \]

### Final Answer:
\[ \lim_{x \to 0} \left( \frac{\sin(\text{sgn}(x))}{\text{sgn}(x)} \right) = 0 \]