I found a question that asked to find the limiting value of $$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{…}}}}}$$If you make the substitution $x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{…}}}}}$ it simplifies to $x=10\sqrt{x}$ which has solutions $x=0,100$. I don’t understand how $x=0$ is a possible solution, I know that squaring equations can introduce new, invalid solutions to equations and so you should check the solutions in the original (unsquared) equation, but doing that here doesn’t lead to any non-real solutions or contradictions. I was wondering if anyone knows how $x=0$ turns out as a valid solution, is there an algebaric or geometric interpretation? Or is it just a “special case” equation?

A similar question says to find the limiting value of $\sqrt{6+5\sqrt{6+5\sqrt{6+5\sqrt{…}}}}$, and making a similar substituion for $x$ leads to

$$x=\sqrt{6+5x}$$

$$x^2=6+5x$$

which has solutions $x=-1,6$. In this case though, you could substitute $x=-1$ into the first equation, leading to the contradiction $-1=1$ so you could satisfactorily disclude it.Is there any similar reasoning for the first question? I know this might be a stupid question but I’m genuinely curious 🙂

**Answer**

Denote the given problem as $x$, then

\begin{align}

x&=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}\\

&=10\cdot10^{\large\frac{1}{2}}\cdot10^{\large\frac{1}{4}}\cdot10^{\large\frac{1}{8}}\cdot10^{\large\frac{1}{16}}\cdots\\

&=10^{\large1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}\\

&=10^{\large y}

\end{align}

where $y$ is an infinite geometric series in which its value is

\begin{align}

y

&=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\\

&=\frac{1}{1-\frac{1}{2}}\\

&=2

\end{align}

Therefore

\begin{equation}

x=10^{\large 2}=100

\end{equation}

**Attribution***Source : Link , Question Author : Anthony Muleta , Answer Author : Anastasiya-Romanova 秀*