Recurring Nested Radical of $0$ I'm having difficulty in determining the value of a recurring nested radical where $n = 0$. I am using the equations
$x = \dfrac{1 + \sqrt{1 + 4n}}{2}$
and
$n = \dfrac{((2x-1)^2 -1)}{4}$ 
to calculate the value of $x$ when $n =0$
Those equations show that $$\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+...}}}}}=1$$ even though it is my understanding that the value of a recurring nested radical who's value is $0$ should also be $0$.
Thanks
 A: You are correct that $\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+...}}}}}=0$  The equations you cite come from trying to evaluate $\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+...}}}}}$. Normally we do not permit infinite expressions, so we need to find some way to interpret this as a limit of finite expressions.  One way is to say $a_1=\sqrt n, a_2=\sqrt{n+\sqrt n}=\sqrt {n+a_1}, a_{k+1}=\sqrt {n+a_k}$ and take the limit as $k \to \infty$  If the limit exists, call it $x$.  We know that $a_k$ and $a_{k+1}$ must both approach it, so we can write $x=\sqrt{n+x}$ or $x^2=n+x$, giving $x=\frac {1 \pm \sqrt {1+4n}}2$.  You lost the minus sign as an option in the solution.  
We still have not shown that a solution exists, nor which sign we should choose.  For $n=0$ it is easy to see that all the $a_k=0,$ so the limit is $0$, which corresponds to the minus sign and all is well.  For $n \gt 0$, we can see that the $a_k$ are monotonically increasing and that $a_1=\sqrt n \gt 0$  If there is a limit we must use the plus sign.  We can show there is a limit if we can show that the sequence is bounded above.  You should be able to show that if $a_k \lt x$, then $a_{k+1} \lt x$, so this $x$ is an upper bound, the limit exists, and it is $x$
