VMO

We can see that $x_1=1=2\sin \frac{\pi }{6}$, $y_1=\sqrt{3}=2\cos \frac{\pi }{6}$. So we will prove, by induction, that for all positive integers $n$, we have $$x_n=2\sin \frac{\pi }{3\cdot 2^n},y_n=2\cos \frac{\pi }{3\cdot 2^n}.(1)$$ Indeed, for $n=1$ the statement is true. Assume that (1) is true for $n$. Applying the recurrent relation, we get $$\begin{array}{rl}{x}_{n+1}& =\sqrt{2-y_n}=\sqrt{2-2\cos \frac{\pi }{3\cdot 2^n}}\\ & =\sqrt{4{\sin }^2\frac{\pi }{3\cdot 2^{n+1}}}=2\sin \frac{\pi }{3\cdot 2^{n+1}}\end{array}$$ and $$y_{n+1}=\frac{x_n}{x_{n+1}}=\frac{2\sin \frac{\pi }{3\cdot 2^n}}{2\sin \frac{\pi }{3\cdot 2^{n+1}}}=2\cos \frac{\pi }{3\cdot 2^{n+1}}.$$ So (1) is also true for $n+1$. Hence, (1) is true for all positive integers $n$. From this we have $$\lim x_n=\lim \left(2\sin \frac{\pi }{3\cdot 2^n}\right)=(2\sin 0)=0$$ and $$\lim y_n=\lim \left(2\cos \frac{\pi }{3\cdot 2^n}\right)=(2\cos 0)=2.$$ Hence, $(x_n)$, $(y_n)$ are convergence and $\lim x_n=0$, $\lim y_n=2$. $\square$