Question

The value of $\underset{n\to \infty }{\lim }{n}^2\left\{\sqrt{\left(1-\cos \frac{1}{n}\right)\sqrt{\left(1-\cos \frac{1}{n}\right)\sqrt{\left(1-\cos \frac{1}{n}\right)\dots \infty }}}\right\}$ is

• A. 1
• B. 2
• C. 0
• D. 1 / 2

Answer 1

Answer: (D)

Let
$P=\underset{n\to \infty }{\lim }{n}^2\left\{\sqrt{\left(1-\cos \frac{1}{n}\right)\sqrt{\left(1-\cos \frac{1}{n}\right)\sqrt{\left(1-\cos \frac{1}{n}\right)\dots \infty }}}\right\}$
Put $\frac{1}{n}=x$
$\therefore P=\underset{x\to 0}{\lim }\frac{\sqrt{(1-\cos x)\sqrt{(1-\cos x)\sqrt{(1-\cos x)\dots \infty }}}}{x^2}$
$=\underset{x\to 0}{\lim }\frac{(1-\cos x{)}^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots \infty }}{x^2}$
$=\underset{x\to 0}{\lim }\frac{(1-\cos x)}{x^2}\frac{(1+\cos x)}{(1+\cos x)}$
$=\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^2\underset{x\to 0}{\lim }\frac{1}{(1+\cos x)}$
$=(1{)}^2\times \frac{1}{1+1}=\frac{1}{2}$