Statement-1 : `lim_(x to 0-) ("sin"[x])/([x]) = "sin" [x] != 0`, Where [x] is the integral part of x.
Statement-2 : `lim_(x to 0+) ("sin"[x])/([x]) != 0`, where [x] the integral part of x.

Options

A. Statement - 1 isTurue, Statement-2 is True, Statement-2 is a correct explanation for statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is Not a correct explanation for statement-1
C. Statement-1 is True, Statement-2 is False
D. Statement -1 is False, Statement-2 is True

Answer

To solve the problem, we need to evaluate the two statements regarding the limits involving the sine function and the integral part of \( x \) (denoted as \( [x] \)). Let's analyze each statement step by step.

### Statement 1:
Evaluate \( \lim_{x \to 0^-} \frac{\sin[x]}{[x]} \).

1. **Understanding \( [x] \) as \( x \) approaches 0 from the left**:
- As \( x \) approaches 0 from the left (negative side), \( [x] \) (the greatest integer function) will be \( -1 \).
- Therefore, \( [x] = -1 \) when \( x \) is in the interval \( (-1, 0) \).

2. **Substituting into the limit**:
- We can rewrite the limit as:
\[ \lim_{x \to 0^-} \frac{\sin[x]}{[x]} = \lim_{x \to 0^-} \frac{\sin(-1)}{-1} \]

3. **Calculating the limit**:
- Since \( \sin(-1) = -\sin(1) \), we have:
\[ \lim_{x \to 0^-} \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1) \]

4. **Conclusion for Statement 1**:
- Thus, \( \lim_{x \to 0^-} \frac{\sin[x]}{[x]} = \sin(1) \), which is not equal to 0.
- Therefore, Statement 1 is **True**.

### Statement 2:
Evaluate \( \lim_{x \to 0^+} \frac{\sin[x]}{[x]} \).

1. **Understanding \( [x] \) as \( x \) approaches 0 from the right**:
- As \( x \) approaches 0 from the right (positive side), \( [x] \) will be \( 0 \).
- Therefore, \( [x] = 0 \) when \( x \) is in the interval \( (0, 1) \).

2. **Substituting into the limit**:
- We can rewrite the limit as:
\[ \lim_{x \to 0^+} \frac{\sin[x]}{[x]} = \lim_{x \to 0^+} \frac{\sin(0)}{0} \]
- This results in the indeterminate form \( \frac{0}{0} \).

3. **Applying L'Hôpital's Rule**:
- To resolve the indeterminate form, we apply L'Hôpital's Rule:
\[ \lim_{x \to 0^+} \frac{\sin[x]}{[x]} = \lim_{x \to 0^+} \frac{\cos[x]}{1} \]
- Now, we substitute \( x = 0 \):
\[ \lim_{x \to 0^+} \cos[x] = \cos(0) = 1 \]

4. **Conclusion for Statement 2**:
- Thus, \( \lim_{x \to 0^+} \frac{\sin[x]}{[x]} = 1 \), which is not equal to 0.
- Therefore, Statement 2 is **True**.

### Final Conclusion:
- Both statements are true:
- Statement 1 is true because \( \lim_{x \to 0^-} \frac{\sin[x]}{[x]} = \sin(1) \).
- Statement 2 is true because \( \lim_{x \to 0^+} \frac{\sin[x]}{[x]} = 1 \).
- However, Statement 2 does not explain Statement 1.