What Is The Square Root Of Zero?

Let $S$ be the expression, then we have:

$S=\sqrt{(x+S) } \Rightarrow S^2-S-x=0 \Rightarrow (S-\frac{1}{2})^2=x+\frac{1}{4}$

If $S-\frac{1}{2}=-\sqrt{( x+\frac{1}{4})} \Rightarrow S=\frac{1}{2} -\sqrt{( x+\frac{1}{4})}$

$x>0 \Rightarrow \sqrt{( x+\frac{1}{4})} >\frac{1}{2} \Rightarrow S<0$

Clearly the expression is positive so this is a contradiction thus

$S-\frac{1}{2}= \sqrt{( x+\frac{1}{4})}$ for $x>0$ .

As we are approaching a limit we can take this to be true as x goes to 0 as it is true for all $x>0$

So as $x$ goes to $0$ we get: $(S-\frac{1}{2})^2=\frac{1}{4} \Rightarrow S-\frac{1}{2}=\frac{1}{2} \Rightarrow S=1$

I think maybe this finally shows why $S \neq 0$ .

Note: Consider the recurrence relation $a_{n+1}=\sqrt{(a_n+x)}, a_0=\sqrt{x}$ so $a_{\infty}$ is equal to the expression.

Clearly $\sqrt{(x+b)} <\sqrt{(x+c)}$ for $c>b>0$

$a_1>a_0$ . Assume $a_{k} >a{k-1} \Rightarrow a_{k+1}=\sqrt{(x+\sqrt{a_k})}>\sqrt{(x+\sqrt{a_{k-1}})}=a_k$ by the above inequality.

It is proved by induction $a_n$ is increasing and, as for $x<1$ we have $x<\sqrt{x} =a_0<a_{\infty}$

This statement is true as $x$ goes towards 0 which is why for $S=0$ the limit is not 0.

The general computation for limit

$\begin{aligned} \lim_{x\to 0^+} \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{...}}}}}}}}}}&\Rightarrow \lim_{x\to 0^+}\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{...}}}}}}}}}} \qquad{\text{Apply the quadratic equation: } x^2-x-C \space\text{if } \sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{C+\sqrt{...}}}}}}}}}} } \\&\Rightarrow x^2-x-0=0 \\&\Rightarrow x^2-x =0 \\&\Rightarrow x(x-1)=0 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{\text{Use Zero Factor Principle}}\\&\Rightarrow \therefore x=1 \space\space \square \end{aligned}$

$\large\text{ADIOS!!!}$

Let's substitute this nested radical, into a real number p. We can obviously see that P^2-p-x=0 Thus, P=(1+sqrt(1+4x))/2 (The conjugate other p doesn't exist since root is defined only for non-negative numbers). Thus, We can take the limit of p when it goes to 0 from right, and we obviously get it to be 1. (Since the function is continous)