<s>[INST] The limit of a nested radical Let $f(a,x)=\sqrt{a(a-1)+x}$, $f^{(n)}(a,x)$ denotes the $n^{th}$ iteration of $x$, where 
\begin{align}f^{(1)}(a,x)=f(a,x),f^{(n)}(a,x)=f^{(n-1)}(a,f(a,x))\end{align}
Find 
\begin{align}\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-f^{(n)}(a,0)}\end{align}
This is a problem from a discussion in a forum, someone say that when $a=2$, the limit is equal to $\pi$, but I can't find the pattern behind this structure, Thanks for your attention.
PS: Expanding it shows
\begin{align}\lim_{n\to\infty}(2a)^{\frac{n}{2}}\sqrt{a-\sqrt{a(a-1)+\sqrt{a(a-1)+\cdots}}}\end{align}
where there are $n$ square root sign behind the minus sign, it is a $\infty \cdot 0$ problem. 
 [/INST] Steps for solution ... With $a=2$ write $f(2,x) = f(x)$.
For the appropriate values of $\theta$ ...
$$\begin{align}
&2\cos\theta = \sqrt{2+2\cos(2\theta)}
\\
&f^{(n)}(2\cos\theta) = 2\cos\frac{\theta}{2^n}
\\
&f^{(n)}(0) = 2\cos\frac{\pi}{2^{n+1}}
\\
&\sqrt{2-2\cos\theta} = 2 \sin \frac{\theta}{2}
\\
&\sqrt{2-f^{(n)}(0)} = 2\sin\frac{\pi}{2^{n+2}}
\\
&\lim_{n\to\infty} 2^n\cdot 2\sin\frac{\pi}{2^{n+1}} = \frac{\pi}{2}
\end{align}$$
</s>