Existence of limit in a recurrence equation: $\alpha_{n+2}=\frac{\alpha_{n}+(1/\alpha_{n+1})}{2}$ [migrated]

Let be $ \boldsymbol{\alpha_{n+2}=\frac{\alpha_{n}+(1/\alpha_{n+1})}{2}}$ a recurrence equation with known $ \alpha_0$ and $ \alpha_1$ . How do you prove that $ \lim_{n\to\infty}\alpha_n$ exists? Note that no conditions are to be assumed about $ \alpha_0$ and $ \alpha_1$ .

I tried to solve this problem with the usual techniques for the classical sequence $ \alpha_{n+2}=\frac{\alpha_{n+1}+1/\alpha_{n+1}}{2}$ , but I did not get any important. My main problem is the unknown signs of the initial conditions, so the sequence can oscillate around 0, but obviously the limit should be 1 or -1.

On the other hand, if this sequence can be divergent, then what conditions about $ \alpha_0$ and $ \alpha_1$ should be necessary and sufficient in order to ensure that it is convergent?