Question

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

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

Answer: (C)

$0$

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

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

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

$=0\times \left[\frac{2}{2}\times \cos \frac{1}{2}+\frac{1}{2}\times \frac{1}{1}\right]=0$