If `f(x)=e^([cotx])`, where `[y]` represents the greatest integer less than or equal to `y` then

Options

A. `lim_(xto(pi^(+))/2)f(x)=1`
B. `lim_(xto(pi^(+))/2)f(x)=1/e`
C. `lim_(xto(pi^(-))/2f(x)=1/e`
D. `lim_(xto(pi^(-))/2f(x)=1`

Answer

To solve the problem, we need to find the limit of the function \( f(x) = e^{[\cot x]} \) as \( x \) approaches \( \frac{\pi}{2} \) from both sides (left and right). Here, \( [y] \) denotes the greatest integer less than or equal to \( y \).

### Step-by-step Solution:

1. **Understanding the Function**:
The function is defined as \( f(x) = e^{[\cot x]} \). We need to analyze the behavior of \( \cot x \) as \( x \) approaches \( \frac{\pi}{2} \).

2. **Behavior of \( \cot x \)**:
The cotangent function, \( \cot x = \frac{\cos x}{\sin x} \), approaches 0 as \( x \) approaches \( \frac{\pi}{2} \) from the left (i.e., \( \frac{\pi}{2}^- \)) and approaches negative infinity as \( x \) approaches \( \frac{\pi}{2} \) from the right (i.e., \( \frac{\pi}{2}^+ \)).
- As \( x \to \frac{\pi}{2}^- \): \( \cot x \to 0 \)
- As \( x \to \frac{\pi}{2}^+ \): \( \cot x \to -\infty \)

3. **Finding \( [\cot x] \)**:
- For \( x \to \frac{\pi}{2}^- \): Since \( \cot x \to 0 \), we have \( [\cot x] = [0] = 0 \).
- For \( x \to \frac{\pi}{2}^+ \): Since \( \cot x \to -\infty \), the greatest integer function will yield \( [\cot x] = -1 \) (as it is the greatest integer less than or equal to a negative value approaching negative infinity).

4. **Calculating the Limits**:
- For \( x \to \frac{\pi}{2}^- \):
\[ f(x) = e^{[\cot x]} = e^{0} = 1 \]
- For \( x \to \frac{\pi}{2}^+ \):
\[ f(x) = e^{[\cot x]} = e^{-1} = \frac{1}{e} \]

5. **Conclusion**:
- Therefore, we have:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = 1 \]
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \frac{1}{e} \]

### Final Answer:
- The limits are:
- \( \lim_{x \to \frac{\pi}{2}^-} f(x) = 1 \)
- \( \lim_{x \to \frac{\pi}{2}^+} f(x) = \frac{1}{e} \)