Let `f(x)= lim_(n->oo)(sinx)^(2n)`

Options

A. Discontinuous at infinite number of points
B. discontinuous at `x=(pi)/(2)`
C. Discontinuous at `x=-(pi)/(2)`
D. none of these

Answer

To solve the problem \( f(x) = \lim_{n \to \infty} (\sin x)^{2n} \), we will analyze the limit based on the value of \( \sin x \).

### Step-by-Step Solution:

1. **Understanding the Limit**:
We start with the function:
\[ f(x) = \lim_{n \to \infty} (\sin x)^{2n} \]
The behavior of this limit depends on the value of \( \sin x \).

2. **Case Analysis**:
We will consider three cases based on the value of \( \sin x \):

- **Case 1**: \( \sin x = 1 \)
- If \( \sin x = 1 \) (which occurs at \( x = \frac{\pi}{2} + 2k\pi \), where \( k \) is an integer), then:
\[ f(x) = \lim_{n \to \infty} 1^{2n} = 1 \]

- **Case 2**: \( \sin x = 0 \)
- If \( \sin x = 0 \) (which occurs at \( x = k\pi \), where \( k \) is an integer), then:
\[ f(x) = \lim_{n \to \infty} 0^{2n} = 0 \]

- **Case 3**: \( 0 < \sin x < 1 \)
- If \( 0 < \sin x < 1 \), then \( (\sin x)^{2n} \) approaches 0 as \( n \) approaches infinity:
\[ f(x) = \lim_{n \to \infty} (\sin x)^{2n} = 0 \]

3. **Conclusion**:
From the analysis above, we can summarize:
- If \( \sin x = 1 \), then \( f(x) = 1 \).
- If \( \sin x = 0 \) or \( 0 < \sin x < 1 \), then \( f(x) = 0 \).

Thus, we can conclude:
\[ f(x) = \begin{cases} 1 & \text{if } x = \frac{\pi}{2} + 2k\pi \\ 0 & \text{if } x = k\pi \text{ or } 0 < \sin x < 1 \end{cases} \]