I'd like to prove $\lim\limits\_{n \rightarrow \infty} \underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}\_{n\textrm{ square roots}}=2$ using Banach's Fixed Point theorem. 

I think I should use the function $f(x)=\sqrt{2+x}$. This way, if I start the iterations for example with $x\_0=0$, I will have $x\_1=\sqrt2$. When I calculate $x\_2$ I will get $\sqrt{2+\sqrt{2}}$. And $x\_3 = \sqrt{2+\sqrt{2+\sqrt{2}}}$ and so on. I can see that these iterations are monotone increasing, but how can I show that this converges to 2?

Pseudo-related formula I found: <http://en.wikipedia.org/wiki/Vieta_formula>

Many thanks in advance!

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Following clark's advice, here's my proof this is a contraction. I'm using the interval $D=[0, 2]$.

$f'(x)=\frac{1}{2\sqrt{x+2}}$, which is monotone decreasing. This means its highest value in $D$ is $0$. $f'(0)=\frac{1}{2\sqrt{2}} < 1$. The rate $M$ of the contraction is then $\frac{1}{2\sqrt{2}}$.