calculus - Where is the lost root?

Suppose we need to solve $x^2=x$. This is simple equation and roots are $x=0,1$.
It is obvious that right hand side must be $\geq 0$ ,so we can write equation as
$$x=\sqrt{x} \tag{1}$$this eq. has the same roots

now put $\sqrt{x} $ in r.h.s instead of $x$ and have
$$x=\sqrt{\sqrt{x}} \tag{2}$$ has the same roots.
again put in r.h.s $x \mapsto \sqrt{x}$so
$$x=\sqrt{\sqrt{\sqrt{x}}} \tag{3}$$ the same roots .
put it over and over again ....
$$x=\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...\sqrt{x}}}}}}\\x=\underbrace{\sqrt[\large{2^n}]{x}}_{n \to \infty} \tag{4}$$ we can write (4)

$$x=\lim_{n\to \infty}(x)^{\dfrac{1}{2^n}}$$ where $x $ is bounded number $\in \mathbb{R}$

so$$\lim_{n\to \infty}(x)^{\dfrac{1}{2^n}}\to 1 \\w.r.t. (4) \to \\x=1$$
$\bf{Question}$: Is the conclusion correct ? If yes ,Where is the other root ?
Thanks in advance .

Answer

$$\lim_{n\to \infty}(x)^{\dfrac{1}{2^n}}= 1$$ only if $x\neq 0.$

In case of $x=0$, we have \begin{align}0&=\lim_{n\to \infty}(0)^{\dfrac{1}{2^n}}\\&=\lim_{n \to \infty}0\\&=0\end{align}

Therefore, $0$ is also a solution.