The value of `lim_(ntooo) (1)/(1+nsin^(2)nx)" can be " (n inN)`

Options

A. `"ln "a_(n)`
B. `e^(a_(1))`
C. `a_(1)`
D. `a_(n)`

Explanation

To solve the limit \( \lim_{n \to \infty} \frac{1}{1 + n \sin^2(nx)} \), we will analyze the behavior of the expression as \( n \) approaches infinity.

### Step-by-Step Solution:

1. **Identify the behavior of \( \sin^2(nx) \)**:
The function \( \sin^2(nx) \) oscillates between 0 and 1 for any value of \( x \). Therefore, we can conclude that:
\[ 0 \leq \sin^2(nx) \leq 1 \]

2. **Consider different cases for \( x \)**:
- **Case 1**: When \( x = m\pi \) (where \( m \) is an integer):
\[ \sin^2(m\pi) = 0 \]
Thus, the limit becomes:
\[ \lim_{n \to \infty} \frac{1}{1 + n \cdot 0} = \frac{1}{1 + 0} = 1 \]

- **Case 2**: When \( x \) is not a multiple of \( \pi \):
In this case, \( \sin^2(nx) \) will oscillate and will not be equal to 0. Therefore, it will take values in the interval \( (0, 1] \). As \( n \) approaches infinity, \( n \sin^2(nx) \) will approach infinity because:
\[ n \sin^2(nx) \to \infty \quad \text{(since } \sin^2(nx) \text{ is bounded away from 0)} \]
Thus, the limit becomes:
\[ \lim_{n \to \infty} \frac{1}{1 + n \sin^2(nx)} = \frac{1}{1 + \infty} = 0 \]

3. **Conclusion**:
Therefore, we can summarize the results:
- If \( x = m\pi \), then the limit is \( 1 \).
- If \( x \) is not a multiple of \( \pi \), then the limit is \( 0 \).

### Final Result:
\[ \lim_{n \to \infty} \frac{1}{1 + n \sin^2(nx)} = \begin{cases} 1 & \text{if } x = m\pi \text{ for } m \in \mathbb{Z} \\ 0 & \text{if } x \text{ is not a multiple of } \pi \end{cases} \]