Show that the limit of the resulting sequence is $\sqrt{2}$. Consider the sequence of rational numbers defined recursively by the following formula: $$(x,y) \mapsto (x^2 + 2y^2, 2xy)$$ and starting at the point $(x,y) = (2,1)$. Show that the limit of the ratio of the resulting sequence is $\sqrt{2}$.
Is there any easy way to look it down and solve it.
I was writting in the form $$(x,y) \mapsto (x^2 + 2y^2, 2xy)= ((x+y)^2+x^2, (x-y)^2+x^2)$$
Still it is not showing any easy calculation later.

As per Yves' response, but a more general approach for $$(x,y) \mapsto (x^2+ay^2, 2xy), a >0$$
we have 
$$\frac{x^2+ay^2}{2xy}= \frac{1}{2}\left(\frac{x}{y}+a\frac{y}{x}\right)$$
which is equivalent to
$$s_{n+1}=\frac{1}{2}\left(s_{n}+\frac{a}{s_{n}}\right)$$
by noting $s_n=\frac{x_n}{y_n}$, with a proof here of the fact that this sequence has $\sqrt{a}$ as the limit.
