`underset(xto(pi)/(2))(lim)([(x)/(2)])/(log_(e)(sinx))([.]` denotes greatest integer function)

Options

A. 0
B. 1
C. -1
D. Doesn't exist

Answer

To solve the limit problem \( \lim_{x \to \frac{\pi}{2}} \frac{\lfloor \frac{x}{2} \rfloor}{\log_e(\sin x)} \), we will analyze the left-hand limit (LHL) and the right-hand limit (RHL) as \( x \) approaches \( \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Identify the Limits**:
We need to find both the left-hand limit (LHL) and the right-hand limit (RHL) as \( x \) approaches \( \frac{\pi}{2} \).

2. **Left-Hand Limit (LHL)**:
\[ \text{LHL} = \lim_{x \to \frac{\pi}{2}^-} \frac{\lfloor \frac{x}{2} \rfloor}{\log_e(\sin x)} \]
As \( x \) approaches \( \frac{\pi}{2} \) from the left, \( \lfloor \frac{x}{2} \rfloor \) approaches \( \lfloor \frac{\frac{\pi}{2}}{2} \rfloor = \lfloor \frac{\pi}{4} \rfloor = 0 \) (since \( \frac{\pi}{4} \approx 0.785 \)).

For \( \log_e(\sin x) \):
\[ \sin x \to \sin\left(\frac{\pi}{2}\right) = 1 \quad \Rightarrow \quad \log_e(\sin x) \to \log_e(1) = 0 \]

Thus, we have:
\[ \text{LHL} = \frac{0}{0} \quad \text{(indeterminate form)} \]

3. **Evaluate the LHL Using Substitution**:
Let \( x = \frac{\pi}{2} - h \) where \( h \to 0^+ \):
\[ \text{LHL} = \lim_{h \to 0^+} \frac{\lfloor \frac{\frac{\pi}{2} - h}{2} \rfloor}{\log_e(\sin(\frac{\pi}{2} - h))} \]
\[ = \lim_{h \to 0^+} \frac{\lfloor \frac{\pi}{4} - \frac{h}{2} \rfloor}{\log_e(\cos h)} \]
Since \( \lfloor \frac{\pi}{4} - \frac{h}{2} \rfloor \) approaches \( 0 \) as \( h \to 0 \), and \( \log_e(\cos h) \to \log_e(1) = 0 \):
\[ \text{LHL} = \frac{0}{\log_e(\cos h)} \to 0 \]

4. **Right-Hand Limit (RHL)**:
\[ \text{RHL} = \lim_{x \to \frac{\pi}{2}^+} \frac{\lfloor \frac{x}{2} \rfloor}{\log_e(\sin x)} \]
As \( x \) approaches \( \frac{\pi}{2} \) from the right, \( \lfloor \frac{x}{2} \rfloor \) still approaches \( \lfloor \frac{\pi}{4} \rfloor = 0 \).

For \( \log_e(\sin x) \):
\[ \sin x \to 1 \quad \Rightarrow \quad \log_e(\sin x) \to 0 \]

Thus, we have:
\[ \text{RHL} = \frac{0}{0} \quad \text{(indeterminate form)} \]

5. **Evaluate the RHL Using Substitution**:
Let \( x = \frac{\pi}{2} + h \) where \( h \to 0^+ \):
\[ \text{RHL} = \lim_{h \to 0^+} \frac{\lfloor \frac{\frac{\pi}{2} + h}{2} \rfloor}{\log_e(\sin(\frac{\pi}{2} + h))} \]
\[ = \lim_{h \to 0^+} \frac{\lfloor \frac{\pi}{4} + \frac{h}{2} \rfloor}{\log_e(\cos h)} \]
Again, \( \lfloor \frac{\pi}{4} + \frac{h}{2} \rfloor \) approaches \( 0 \) and \( \log_e(\cos h) \to 0 \):
\[ \text{RHL} = \frac{0}{\log_e(\cos h)} \to 0 \]

6. **Conclusion**:
Since both the LHL and RHL are equal:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\lfloor \frac{x}{2} \rfloor}{\log_e(\sin x)} = 0 \]

### Final Answer:
\[ \boxed{0} \]