proof problem on proving $x_{1}=1,x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)$ has a limit Sequence $\{x_{n}\}$ is defined by $$x_{1}=1,x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)$$, as for proving whether the sequence  has a limit, one of my friends told me his proof as following:

First, assuming the sequence has a limit $x$, then take limit on both
  sides of $$x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)$$,
  that is $$\lim _{x\rightarrow \infty }x_{n+1} =\lim _{x\rightarrow\infty }(\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right))$$, which can
  be simplified to$$x =\frac{1}{2}\left(x+\frac{2}{x}\right)$$,which is
  easily solved to show that $x=\sqrt 2 $ or $  -\sqrt2$, and it is easy to show the limit should be $\sqrt 2 $,since we have already  worked out the value of the limit, so also proved the existence of the limit as a side effect.



*

*Does the proof really prove that the sequence has a limit? 

*Is it possible that we computed out the limit value of a sequence  using similar method as above(without proving the existence of the limit before) , but later found that the sequence doesn't have a limit ?

For $x>0$ we have  $x+\frac2x\geq 2\sqrt{x} \sqrt{\frac2x}=2\sqrt2$. So $x_n\ge \sqrt2$. Further $\frac{x_{n+1}}{x_n}=\frac12\left(1+\frac2{x_n^2}\right)\le 1.$
So the sequence converges. Its limit is already known.
