`lim_(xrarr oo) (n^p sin^2(n!))/(n+1),0ltplt1`, is equal to

Options

A. 0
B. `oo`
C. 1
D. none of these

Explanation

To solve the limit \( \lim_{n \to \infty} \frac{n^p \sin^2(n!)}{n + 1} \) where \( 0 < p < 1 \), we can follow these steps:

### Step 1: Rewrite the Limit
We start with the limit expression:
\[ \lim_{n \to \infty} \frac{n^p \sin^2(n!)}{n + 1} \]
We can simplify the denominator by dividing both the numerator and the denominator by \( n \):
\[ = \lim_{n \to \infty} \frac{n^p \sin^2(n!)}{n(1 + \frac{1}{n})} \]

### Step 2: Simplify the Expression
This gives us:
\[ = \lim_{n \to \infty} \frac{n^{p-1} \sin^2(n!)}{1 + \frac{1}{n}} \]
Here, we notice that as \( n \to \infty \), \( 1 + \frac{1}{n} \) approaches 1. So we can focus on the numerator:
\[ = \lim_{n \to \infty} n^{p-1} \sin^2(n!) \]

### Step 3: Analyze the Behavior of \( \sin^2(n!) \)
The term \( \sin^2(n!) \) oscillates between 0 and 1 for all \( n \). Therefore, we can say:
\[ 0 \leq \sin^2(n!) \leq 1 \]
This means that:
\[ 0 \leq n^{p-1} \sin^2(n!) \leq n^{p-1} \]

### Step 4: Determine the Limit of \( n^{p-1} \)
Since \( p < 1 \), we have \( p - 1 < 0 \). Thus, as \( n \to \infty \):
\[ n^{p-1} \to 0 \]
This implies that:
\[ \lim_{n \to \infty} n^{p-1} \sin^2(n!) = 0 \]

### Step 5: Conclude the Limit
By the squeeze theorem, since \( n^{p-1} \sin^2(n!) \) is squeezed between 0 and \( n^{p-1} \) (which approaches 0), we conclude that:
\[ \lim_{n \to \infty} \frac{n^p \sin^2(n!)}{n + 1} = 0 \]

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