Pi Appoximation: Simpler Solution to Limit? May be a ridiculous question, but I wanted to see if MSE had "simpler" proofs for Viete's approximation (specifically, using an equation derived from Viete's formula) of $\pi$: $$\lim_{x \to \infty}2^x \left(\sqrt{2-\sqrt{2+\sqrt{2 + \sqrt{2 +\sqrt(2)+...} }}}\right) = \pi$$ for x square roots (and only one minus sign; rest are addition).
By "simpler," I'm sort of looking for a proof not using Viete's formulas at the least, but I would also like to know if it's possible to show this with more basic limit methods (e.g., Taylor Series(?), taking natural log, etc)
Thank you kindly!

You can use the fact that 
$$\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}=2\cos\frac{\pi}{2^{n+1}},$$
where $n$ is the number of radicals in LHS. Thus your sequence is
$$2^n\sqrt{2-2\cos\frac{\pi}{2^n}}=2^{n+1}\sin\frac{\pi}{2^{n+1}}\to \pi$$
as $n\to\infty$, since $$\lim_{x\to 0}\frac{\sin\pi x}{x}=\pi.$$
