Convergence of $x_{n+1} = \frac12(x_n + \frac2{x_n}).$ [duplicate]

Let $x_1=1$ and $$x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).$$ Prove or disprove $(x_n)$ is convergent and show the limit.

When I tried working on it I found the sequence was bounded by square root of 2 and it is was monotone. But apparently the sequence is not bounded by square root of two and is not monotone. But I have no idea why. Any help would be greatly appreciated! Thanks!

Let $y_n = x_{n}^2$, then \begin{align*} \\y_{n+1} = x_{n+1}^2 &= \frac{1}{4}\left(x_n + \frac2{x_n}\right)^2 \\ &=\left(\frac{x_n}{2}\right)^2 +\left(\frac{1}{x_n}\right)^2 + 1 \ge 2\sqrt{\frac{x_n}{2}\cdot\frac{1}{x_n}} +1 = \sqrt{2}+1 \gt 2 \tag{1} \end{align*} Thus by (1), we have $x_n \gt \sqrt{2}~~(n\ge 2)$. Also, by the given condition, we have: $$x_{n+1}-x_n=\frac{1}{2}x_n+\frac{1}{x_n}-x_n=\frac{1}{x_n}-\frac{x_n}{2} \tag{2}$$ (2) implies that the sequence is decreasing when $x_n \gt \sqrt{2}$ and we have already known that for $n \ge 2$, $x_n \gt \sqrt{2}$. So $x_n$ is decrease and bound below by $\sqrt{2}$. Hence, $x_n$ is convergent.

Note: Let $f(x) = \frac{1}{x} +\frac{x}{2}$, then $$f(x)\ge 0 \space\text{when } x \in (-\infty,-\sqrt2] \cup (0,\sqrt2]$$ $$f(x)\lt 0 \space\text{when } x \in (-\sqrt2,0) \cup (\sqrt2, +\infty)$$

$$x_{n+1}-x_n=\frac{1}{2}x_n+\frac{1}{x_n}-x_n=\frac{1}{x_n}-\frac{x_n}{2}$$

$$\text{$(x_n)$bounded above by $\sqrt{2}$ iff $(x_n)$ is monotonically increasing??}$$

$$\text{Does this reduce your problem to check only for one of monotonicity/ boundedness?}$$