The value of `lim_(ntooo) [(2n)/(2n^(2)-1)"cos"(n+1)/(2n-1)-(n)/(1-2n).(n)/(n^(2)+1)]` is

Options

A. 5
B. 6
C. 7
D. none of these

Explanation

To solve the limit

\[ \lim_{n \to \infty} \left[ \frac{2n}{2n^2 - 1} \cos\left(n + 1\right) \cdot \frac{1}{2n - 1} - \frac{n}{1 - 2n} \cdot \frac{n}{n^2 + 1} \right], \]

we will break it down step by step.

### Step 1: Simplify the expression

We start with the limit:

\[ \lim_{n \to \infty} \left[ \frac{2n \cos(n + 1)}{(2n^2 - 1)(2n - 1)} - \frac{n^2}{(1 - 2n)(n^2 + 1)} \right]. \]

### Step 2: Analyze the first term

For the first term, we can rewrite it as:

\[ \frac{2n \cos(n + 1)}{(2n^2 - 1)(2n - 1)}. \]

As \( n \to \infty \), \( \cos(n + 1) \) oscillates between -1 and 1. Therefore, we can focus on the denominator:

\[ (2n^2 - 1)(2n - 1) \approx 4n^3 \text{ as } n \to \infty. \]

Thus, the first term behaves like:

\[ \frac{2n \cdot \text{bounded}}{4n^3} = \frac{\text{bounded}}{2n^2} \to 0 \text{ as } n \to \infty. \]

### Step 3: Analyze the second term

Now consider the second term:

\[ -\frac{n^2}{(1 - 2n)(n^2 + 1)}. \]

As \( n \to \infty \), we can simplify the denominator:

\[ (1 - 2n)(n^2 + 1) \approx -2n^3 \text{ as } n \to \infty. \]

Thus, the second term behaves like:

\[ -\frac{n^2}{-2n^3} = \frac{1}{2n} \to 0 \text{ as } n \to \infty. \]

### Step 4: Combine the results

Now, we combine the results of both terms:

\[ \lim_{n \to \infty} \left[ 0 - 0 \right] = 0. \]

### Final Answer

Thus, the value of the limit is:

\[ \boxed{0}. \]