Question

The value of $\underset{n\to \infty }{\text{lim}}[\frac{2n}{2n^2-1}\cos \frac{n+1}{2n-1}-\frac{n}{1-2n}\cdot \frac{n{(-1)}^n}{n^2+1}]$ is

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

Given $\underset{n\to \infty }{\text{lim}}[\frac{2n}{2n^2-1}\cos \frac{n+1}{2n-1}-\frac{n}{1-2n}\cdot \frac{n{(-1)}^n}{n^2+1}]$
$=\underset{n\to \infty }{\text{lim}}[\frac{2}{2-\frac{1}{n^2}}\cdot \frac{1}{n}\cos \left(\frac{1+\frac{1}{n}}{2-\frac{1}{n}}\right)-\frac{1}{(\frac{1}{n}-2)}\cdot \frac{{(-1)}^n}{(1+\frac{1}{n^2})}\cdot \frac{1}{n}]$

$=\underset{n\to \infty }{\text{lim}}\frac{1}{n}[\frac{2}{2-\frac{1}{n^2}}\cdot \cos \left(\frac{1+\frac{1}{n}}{2-\frac{1}{n}}\right)-\frac{1}{(\frac{1}{n}-2)}\cdot \frac{{(-1)}^n}{(1+\frac{1}{n^2})}]$

$=0\times [\frac{2}{2}\times \cos \frac{1}{2}-\frac{1}{(-2)}\times \frac{+1\text{or}-1}{1}]=0.$