Finding the value of x, $x=\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+....} } } } $

Question

I want to find the value of x where x is given by the following: $$x=\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+....} } } } $$ Noting that the above can be written in the following form: $x=\sqrt{0+x} $, we have $x=0$ or $x=1$
It seems that $x=0$ and $x\ne 1$, however I don't know how to reject the possibility that $x\ne 1$. Any help is appreciated.

Edit: From the comments, I have understood that the question lacks details. This is because the above nested radical can be produced either by the sequence $a_1=0, a_{n+1}=\sqrt{0+a_n}$ or by $a_1=c \gt 0, a_{n+1}=\sqrt{0+a_n}$, which converge to 0 and 1 respectively. And therefore, both the answers x=0 and 1 are correct.

Answer 1

If you want a formal proof, use mathematical induction on $n=$ the number of square root radicals in each term of the sequence.

• At $n=1$ you clearly have $\sqrt0=0$.

• Assume you have $0$ for $n=k$. Then the next term with $n=k+1$ is $\sqrt{0+0}$ thus proving the propagation of the $0$ value for all terms.