Limit of $\lim_{x\rightarrow 1}\sqrt{x-\sqrt[3]{x-\sqrt[4]{x-\sqrt[5]{x-.....}}}}$ I known how to find the limit of my last question which was Finding $\lim_{x\rightarrow 1}\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x....}}}}$, but I couldn't how to start to find the limit of $$\lim_{x\rightarrow 1}\sqrt{x-\sqrt[3]{x-\sqrt[4]{x-\sqrt[5]{x-.....}}}}$$     so I used  a program written in visual basic 6 to calculate it.I found the limit about equal to $29/30$ but I am not sure if this value is right because the accuracy of this language not enough to give me a trust value. Anyhow,I want to know how to find the limit analytically.
 A: It is not proved that a limit exists. 
The figure below will give a new view on this question of limit or no limit. We consider an infinite set of functions $y_1(x)$ , $y_2(x)$ , ... , $y_n(x)$ , ... where the index $n$ is the number of radicals :

In case of the functions with odd index, the limit for $x$ tending to $1$ is  $y(1)=1$  
In case of the functions with even index, the limit for $x$ tending to $1$ is  $y(1)=0$
When the number of radicals tends to infinity without specified parity, the limit is not defined. So we could say that there is no limit for the expression given in the wording of the question.
But, what is interresting is not the set of functions and corresponding curves, but what is the boundary curve of the infinite set of curves. This curve is not included into the set itself. The corresponding function might be on the form $y_{boundary}(x)=f(x)H(x-1)$ where H is the symbol of the Heaviside step function. Looking for $f(x)$ is something else: This is a challenge. 
IN ADDITION :
We can express the firsts terms of the series expansion around $x=1$.  
Let $x=1+\epsilon$ ,  with $\epsilon$ close to $0$.
$y_1=1+\frac{\epsilon}{2}+O(\epsilon^2)$
$y_2=(\frac{2\epsilon}{3})^{1/2}+O(\epsilon^{2/2})$
$y_3=1-(\frac{3\epsilon}{32})^{1/3}+O(\epsilon^{2 /3})$
$y_4=(\frac{4\epsilon}{405})^{1/8}+O(\epsilon^{2/8})$
$y_5=1-(\frac{5\epsilon}{201326592})^{1/15}+O(\epsilon^{2/15})$
$y_6=(\frac{2\epsilon}{ 10296910184203125})^{1/48}+O(\epsilon^{2/48})$
etc.
For $\epsilon=0$, this confirms that $y_n$  is alternatively equal to $0$ and to $1$ when $n$ tends to infinity.
When $n$ tends to infinity, the curves corresponding to the functions $y_n(x)$ tend asymptotically to a curve $f(x)$ for $x>1$. For $x=1$, $y_n(1)$ tends to $1$ if $n$ is odd, or tends to $0$ if $n$ is even. 
The bounding curve corresponding to $f(x)$ is drawn in red on the figures below, represented in various enlargements. Of course, the curve in red is the same for all the figures. It looks like a straight line if we consider a range of $x$ close to $x=1$, but it is a misleading impression : $f(x)$ is not a linear function. The computation for drawing $f(x)$ was made with an algorithm similar to that given by Han de Bruijn. 




