Question

The value of $\displaystyle\lim _{n \rightarrow \infty} n^{2}\left\{\sqrt{\left(1-\cos \frac{1}{n}\right) \sqrt{\left(1-\cos \frac{1}{n}\right) \sqrt{\left(1-\cos \frac{1}{n}\right) \ldots \infty}}}\right\}$ is

• A. 1
• B. 2
• C. 0
• D. 1 / 2

Answer 1

Answer: (D)

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