<s>[INST] Recurrence relation: $x_{n+1} = \frac 12 x_n + \frac 1 {x_n},$ $$x_{n+1} = \frac 12 x_n + \frac 1 {x_n}, x_0 \neq 0$$
$$ a = \frac a2  + \frac 1a \Rightarrow a = \frac {a^2 + 2} {2a} \Rightarrow 2a^2 = a^2 + 2 \Rightarrow a^2 = 2 \Rightarrow a = \pm \sqrt 2$$
If $x_0 > 0$, then any subsequent terms will be positive too. If $x_0 < 0$ , then - negative. Hence, if $x_0 > 0$, then $\lim_{n \rightarrow \infty} x_n = \sqrt 2$. If $x_0 < 0$, then $\lim_{n \rightarrow \infty} x_n = - \sqrt 2$
Is it complete solution? I know that it's correct answer, but may be I missed something in proof?
 [/INST] You can show that the sequence indeed converges by the following simple way:
From the recursion relation we get, $$\left(\frac{x_{n+1}+\sqrt{2}}{x_{n+1}-\sqrt{2}}\right)=\left(\frac{x_{n}+\sqrt{2}}{x_{n}-\sqrt{2}}\right)^2, \forall n\geq1 $$ So iteratively, we get, $$\left(\frac{x_{n+1}+\sqrt{2}}{x_{n+1}-\sqrt{2}}\right)=\left(\frac{x_{0}+\sqrt{2}}{x_{0}-\sqrt{2}}\right)^{2^{n+1}}$$ Rearranging, we get, $$x_{n+1}=\sqrt{2}\frac{(x_0+\sqrt{2})^{2^{n+1}}+(x_0-\sqrt{2})^{2^{n+1}}}{(x_0+\sqrt{2})^{2^{n+1}}-(x_0-\sqrt{2})^{2^{n+1}}}$$ Now if $x_0\ne \sqrt{2},\ x_0>0$, then $|\frac{x_0-\sqrt{2}}{x_0+\sqrt{2}}|<1$.  Hence $$\lim_{n\rightarrow \infty}x_{n+1}=\sqrt{2}$$
If if $x_0\ne -\sqrt{2},\ x_0\leq0$, then $|\frac{x_0+\sqrt{2}}{x_0-\sqrt{2}}|<1$. Hence $$\lim_{n\rightarrow \infty}x_{n+1}=-\sqrt{2}$$ Clearly, if $x_0={}_{-}^+\sqrt{2}$ then 
$$\lim_{n\rightarrow \infty}x_{n+1}={}_{-}^+\sqrt{2}$$ respec. 
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