Question

$\underset{n\to \infty }{lim}{n}^2\left(\sqrt{(1-cos\frac{1}{n})\sqrt{(1-cos\frac{1}{n})\sqrt{(1-cos\frac{1}{n}).....\infty }}}\right)$
is equalte

• A. $0$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

Putting $\frac{1}{n}=x$ Required limit becomes,
$=\underset{x\to 0}{lim}\frac{\sqrt{(1-cosx)\sqrt{(1-cosx)\sqrt{(1-cosx).....\infty }}}}{x^2}$
$=\underset{x\to 0}{lim}\frac{(1-cosx{)}^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^2}+....\infty }}{x^2}$
$=\underset{x\to 0}{lim}\frac{(1-cosx)}{x^2}\times \frac{(1+cosx)}{(1+cosx)}$
$=\underset{x\to 0}{lim}{\left(\frac{sinx}{x}\right)}^2.\underset{x\to 0}{lim}\frac{1}{(1+cosx)}=(1{)}^2.\frac{1}{2}=\frac{1}{2}$