`lim_( n rarr oo) sin [ pi sqrt(n^(2)+1)]` is equal to

Options

A. `oo`
B. 0
C. Does not exist
D. None of these

Answer

To solve the limit \( \lim_{n \to \infty} \sin \left( \pi \sqrt{n^2 + 1} \right) \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the Argument of the Sine Function**:
We start with the limit:
\[ \lim_{n \to \infty} \sin \left( \pi \sqrt{n^2 + 1} \right) \]
We can manipulate the expression inside the sine function:
\[ \sqrt{n^2 + 1} = \sqrt{n^2(1 + \frac{1}{n^2})} = n\sqrt{1 + \frac{1}{n^2}} \]
Therefore, we can rewrite the limit as:
\[ \lim_{n \to \infty} \sin \left( \pi n \sqrt{1 + \frac{1}{n^2}} \right) \]

2. **Simplify the Expression**:
As \( n \to \infty \), \( \sqrt{1 + \frac{1}{n^2}} \) approaches 1. Thus:
\[ \lim_{n \to \infty} \sin \left( \pi n \cdot 1 \right) = \lim_{n \to \infty} \sin(\pi n) \]
The sine function has a periodicity of \( 2\pi \), and \( \sin(\pi n) = 0 \) for all integer values of \( n \).

3. **Evaluate the Limit**:
Therefore, we conclude:
\[ \lim_{n \to \infty} \sin(\pi n) = 0 \]

Thus, the final result is:
\[ \lim_{n \to \infty} \sin \left( \pi \sqrt{n^2 + 1} \right) = 0 \]

### Final Answer:
The limit is \( 0 \).