You have the sequence $x_{n+1} = \sqrt{x_n}$
For $0 \lt x_1 \lt 1$, we have that $x_{n+1} \gt x_n$ and $x_n \lt 1$. So this is a monotonically increasing sequence bounded above, and so has a limit.
The limit $L = \lim_{n \to \infty} x_n$ satisfies $0 \lt x_1 \le L \le 1$ and $L = \sqrt{L}$, thus $L = 1$.
If $x_1 \gt 1$, you get a monotonically decreasing sequence which is bounded below and you get $\lim_{n \to \infty} x_n = 1$, similarly.
Note that, if you don't start with $1$ (or $0$), in a finite number of steps, you will get very close to $1$, but don't really attain $1$. The result of $1$ you see is likely due to the precision limitations of your calculator.