`lim_(ntooo)sin[pisqrt(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 1: Simplify the expression inside the sine function
We start with the expression inside the sine function:
\[ \sqrt{n^2 + 1} \]
As \( n \) approaches infinity, we can approximate \( \sqrt{n^2 + 1} \) using the following simplification:
\[ \sqrt{n^2 + 1} = \sqrt{n^2(1 + \frac{1}{n^2})} = n\sqrt{1 + \frac{1}{n^2}} \approx n\left(1 + \frac{1}{2n^2}\right) = n + \frac{1}{2n} \quad \text{(using Taylor expansion)} \]

### Step 2: Substitute back into the limit
Now substituting this back into our limit, we get:
\[ \lim_{n \to \infty} \sin\left(\pi\left(n + \frac{1}{2n}\right)\right) \]

### Step 3: Rewrite the sine function
Using the property of sine, we can rewrite the expression:
\[ \sin\left(\pi n + \frac{\pi}{2n}\right) \]
Using the sine addition formula, we know that:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
So we can write:
\[ \sin\left(\pi n + \frac{\pi}{2n}\right) = \sin(\pi n)\cos\left(\frac{\pi}{2n}\right) + \cos(\pi n)\sin\left(\frac{\pi}{2n}\right) \]

### Step 4: Evaluate the components
1. **Evaluate \( \sin(\pi n) \)**: Since \( n \) is an integer, \( \sin(\pi n) = 0 \).
2. **Evaluate \( \cos(\pi n) \)**: Since \( n \) is an integer, \( \cos(\pi n) = (-1)^n \).
3. **Evaluate \( \sin\left(\frac{\pi}{2n}\right) \)**: As \( n \to \infty\), \( \frac{\pi}{2n} \to 0 \) and thus \( \sin\left(\frac{\pi}{2n}\right) \approx \frac{\pi}{2n} \).

### Step 5: Combine results
Putting this all together, we have:
\[ \sin\left(\pi n + \frac{\pi}{2n}\right) = 0 \cdot \cos\left(\frac{\pi}{2n}\right) + (-1)^n \cdot \frac{\pi}{2n} \]
This simplifies to:
\[ (-1)^n \cdot \frac{\pi}{2n} \]

### Step 6: Take the limit
Now, we take the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} (-1)^n \cdot \frac{\pi}{2n} = 0 \]
Since \( \frac{\pi}{2n} \to 0 \) as \( n \to \infty \).

### Final Answer
Thus, the limit is:
\[ \boxed{0} \]