This is a nice limit $$\lim_{n\rightarrow \infty}\underbrace{\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}}_{n\text{ times}}$$ and it is solved with well-known trigonometry formulas. The result is 2. The question is: can be, this limit, solved in other ways?
Thanks.
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If the limit exists, the recursive function is very simply: $$n=\sqrt{2+n}$$
Hence:
$$n^2-n-2=0 \\ (n-2)(n+1)=0 \\n=2$$
$n$ is indeed $2$.
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Seeing as there is no $n$ in your expression, having $\lim_{n \to \infty}$ doesn't really make sense. I would just express it as $\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}$
Just suppose there is some number $A$ such that $A = \sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}$. then we can square and subtract two from both sides and get $A^2 -2 = \sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}$. But the right-hand side is the same in both of these, so we have $A^2 - 2 = A$. Solve this quadratic for $A$ and we see that in fact $A = 2$
If $x_0=\sqrt 2$ and $x_{n+1}=\sqrt{2+x_n}$, then you can prove by induction that $x_{n}<2$ and $x_n$ is increasing. So a limit must exist.
But any limit must be a positive $x$ satisfying $x = \sqrt{2+x}$. So $x=2$.