If `f(x) = sin(lim_(t rarr 0)(2x)/picot^(-1) (x/t^2))`, then `int_(-(pi)/(2))^((pi)/(2))f(x)` dx is equal to (where , `x ne0`)

Options

A. `-2`
B. `-1`
C. 0
D. 2

Answer

To solve the problem, we need to evaluate the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx \) where \( f(x) = \sin\left(\lim_{t \to 0} \frac{2x}{\pi \cot^{-1}\left(\frac{x}{t^2}\right)}\right) \).

### Step-by-Step Solution:

1. **Understanding \( \cot^{-1} \)**:
- Recall that \( \cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right) \). For \( y = \frac{x}{t^2} \), we can rewrite it as:
\[ \cot^{-1}\left(\frac{x}{t^2}\right) = \tan^{-1}\left(\frac{t^2}{x}\right) \]

2. **Evaluating the Limit**:
- We need to evaluate \( \lim_{t \to 0} \frac{2x}{\pi \tan^{-1}\left(\frac{t^2}{x}\right)} \).
- As \( t \to 0 \), \( \frac{t^2}{x} \to 0 \) which implies \( \tan^{-1}\left(\frac{t^2}{x}\right) \to 0 \).
- Therefore, we have:
\[ \lim_{t \to 0} \tan^{-1}\left(\frac{t^2}{x}\right) = 0 \]

3. **Substituting the Limit into \( f(x) \)**:
- Thus, we can substitute this limit back into \( f(x) \):
\[ f(x) = \sin\left(\lim_{t \to 0} \frac{2x}{\pi \tan^{-1}\left(\frac{t^2}{x}\right)}\right) = \sin(0) = 0 \]
- This means \( f(x) = 0 \) for all \( x \neq 0 \).

4. **Evaluating the Integral**:
- Now we can evaluate the integral:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 0 \, dx = 0 \]

5. **Conclusion**:
- The value of the integral is:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = 0 \]

### Final Answer:
The value of the integral is \( 0 \).