How to prove that the limit belongs to $(0, 1)$? In a sequence of positive numbers $$a_{1}, a_{2},\dots,a_{n},\dots,$$
each of the terms $a_n$ is equal to either
$\dfrac{ a_{n-1}  }{ 2}$ or $\sqrt{a_{n - 1}}$. 
Can this sequence have a limit belonging to the interval $(0, 1)$?
 A: I assume that you meant to say that for each $n$, we have either $a_n = \frac 12 a_{n-1}$ or $a_n = \sqrt{a_{n-1}}$. 
The answer to your question (as I understand it) is no. Suppose for the purpose of contradiction that $\lim_{n \to \infty} a_n = L$ with $0 < L < 1$.  First, note that there must be infinitely many $n$ such that $a_{n} = \frac 12 a_{n-1}$.  Otherwise, the sequence would converge to a fixed point of the function $f(x) = \sqrt{x}$, namely $0$ or $1$.
Let $N$ be such that $n > N$ implies that $|a_n - L| < L/8$.  Select an $n>N$ such that $a_{n+1} = \frac 12 a_n$.  We note that 
$$
|a_{n+1} - L| \geq |a_{n+1} - a_n| - |a_n - L| = \frac 12 a_n - |a_n - L| \\
 > \frac 12 (L - L/8) - L/8 = \frac 5{16}L > L/8.
$$
However, our definition of $N$ implies that since $n+1 > N$, we have $|a_{n+1} - L| < L/8$.  So, we have a contradiction.
A: Assume that $\lim_{n \to \infty} a_n = a$ exists, with $0 < a < 1$. Then there exists an $N$ such that for all $n \ge N$
$$
 \frac 34 a < a_n < \frac 54 a \, .
$$
It follows that for $n> N$
$$
 \color{red}{a_n} > \frac 34 a > \frac 12 \cdot \frac 54 a \color{red}{> \frac 12 a_{n-1}}
$$
and consequently
$$
 a_n = \sqrt{a_{n-1}} \, .
$$
The limit must therefore satisfy 
$$
 a = \sqrt a 
$$
which is not possible for $0 < a < 1$.
Therefore such a sequence does not exist. 
A: The answer is: No.
Either it holds $a_n \ge 1$ for all $n\in\Bbb{N}$ then obviously $$\lim_{n\to\infty} a_n \ge 1$$ if it exists.
Or there is one $n_0 \in \Bbb{N}$ s.t. $a_n < 1$ but then $\sqrt{a_n -1}$ is not defined so for all following elements of the sequence it has to hold $$a_n =  \dfrac{ a_{n-1}  }{ 2}$$ but then obviously $$\lim_{n\to\infty} a_n = 0$$
