50 follower problem

Let $\alpha = \sqrt {\bigg( 1- \cos(\frac{1}{x}) \bigg) \sqrt {\bigg( 1-\cos(\frac{1}{x}) \bigg) \sqrt{\bigg( 1-\cos(\frac{1}{x}) \bigg) \sqrt {\ldots}}}}$

So, we obtain $\dfrac{\alpha^2}{1-\cos \left(\frac1x \right)} = \alpha$

$\Rightarrow \alpha = 1-\cos(\frac1x)$ .

So we have the limit as:

$\lim_{x \to \infty} x^2 \alpha = \lim_{x \to \infty} x^2 \left( 1-\cos(\frac1x) \right) = \lim_{x \to \infty} x^2 \cdot \left (2\sin^2 \left(\frac{1}{2x} \right) \right)$

$= \lim_{x \to \infty} \dfrac{\sin^2 \left( \frac{1}{2x} \right)}{2 \cdot \left(\frac{1}{2x} \right)^2} =\boxed{ \dfrac12}$

$\sqrt {\bigg( 1-cos\frac{1}{x} \bigg) \sqrt {\bigg( 1-cos\frac{1}{x} \bigg) \sqrt{\bigg( 1-cos\frac{1}{x} \bigg) \sqrt {... }}}} =y$ $\ 1-y = cos\frac{1}{x}$ $\text{so}\lim_{x\to \infty} x^2 \sqrt {\bigg( 1-cos\frac{1}{x} \bigg) \sqrt {\bigg( 1-cos\frac{1}{x} \bigg) \sqrt{\bigg( 1-cos\frac{1}{x} \bigg) \sqrt {... }}}} = \lim_{y\to 0} \bigg(\frac{1}{cos^{-1} (1-y)}\bigg)^2 \times y$ $\cos^{-1}(1-y) = x$ $\lim_{y\to 0} \bigg(\frac{1}{cos^{-1} (1-y)}\bigg)^2 \times y = lim_{z\to 0} \bigg( \bigg( \frac{1}{z}\bigg)^{2} \times (1-cos (z)) \times \frac{1+cos (z)}{1+cos (z)} \bigg) = \frac{1}{2}$ If you like this solution upvote it and reshare the problem if you don't like it post another solution and reshare the problem also :)