Find the value of $\sqrt{10\sqrt{10\sqrt{10...}}}$ 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 :)
 A: $$x= 10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}} \Rightarrow \left(\frac{x}{10}\right)^2 = x$$
$$\Rightarrow x^2 -100x = 0 \Rightarrow x(x-100) = 0$$ Since $x$ is positive (why?), we conclude that $x=100$.         
A: If $$x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$
then it is true that $x=10\sqrt x$. 
It is not true that if $x=10\sqrt x$, then $$x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}.$$
$0$ is indeed one solution, but it is not a valid one.
A: Here the question is (apparently) about the sequence defined recursively by
$$
\begin{aligned}
a_1&=10,\\
a_{n+1}&=10\sqrt{a_n}\quad \text{for all positive integers $n$.}
\end{aligned}
$$
It often happens that students embark on a quest of finding the value of the limit of a sequence before ascertaining that what they try to calculate actually exists. Here we should first satisfy ourselves that $\lim_{n\to\infty}a_n$ actually exists as a real number. A convenient tool for this is the theorem telling us that a bounded increasing sequence converges towards some limit $A$. Here it is easy to prove by induction that $a_{n+1}>a_n$ for all $n$, and also that $a_n<100$ for all $n$.
A part of that theorem then says that $a_1\le A\le 100$. After all, all the members of the sequence are in the closed interval $[10,100]$ so their limit cannot be outside this interval either. The recurrence relation (and continuity of the square root function in this interval) then give us the equation $A=10\sqrt{A}$ allowing us to deduce that either $A=0$ or $A=100$. The former solution, however, does not belong to this interval and can be discarded as an alternative.
Your equation having another solution is just a coincidence. In this case as well as with the recurrence formula $a_{n+1}=\sqrt{6+5a_n}$.

Compare with the following other common misuse of limits:
$$
S=1-1+1-1+1-1+\cdots
$$
Some are want for rewriting this as
$$
S=1-(1-1+1-1+1-\cdots)=1-S,
$$
solving the equation $S=1-S$, and concluding that $S=1/2$.
Where's the mistake? It is in the first line, where the student treats the sum of this series as if it has a value, i.e. as if it converges, by the act of calling it $S$. The moral: Beware of naming things that don't necessarily exist :-)
A: Alternatively, let $a_1,\,a_2,\,a_3,\,\cdots,\,a_n$ be the following sequence
$$10\sqrt{10},\,10\sqrt{10\sqrt{10}},\,10\sqrt{10\sqrt{10\sqrt{10}}},\,\cdots,10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots\sqrt{10}}}}}}_{\large n\,\text{times}}$$
respectively.
Notice that
$$\large a_n=10^{\Large 2-2^{-n}}$$
Hence
$$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}=\large\lim_{n\to\infty}\, a_n=\lim_{n\to\infty}\,10^{\Large 2-2^{-n}}=\bbox[3pt,border:3px #FF69B4 solid]{\color{red}{100}}$$
A: 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}
A: Consider the sequence $a_{n+1}= 10\sqrt{a_n}$ with $a_0 \ge 0$.  The expression you have could be considered the limit as $n \to \infty$, if it exists. 
Further, $a_{n+1} = \sqrt{100 \cdot a_n}$ gives that $a_{n+1}$ is between $100$ and $a_n$.  So if you start with $a_0 > 0$, you have a nice bounded monotone sequence, which converges to a limit and the limit will of course be $100$.  
OTOH if you start with $a_0 = 0$, clearly the limit exists still, and you have the limit as $0$.  
As your expression is ambiguous about what $a_0$ could be, you get both solutions.
A: have you noticed that $10$ is no special value. In this replace $10$ by $a$ i.e. any real number, equation will be still $x=a\sqrt{x}$, and $0$ will be a solution, so this can be regarded as a defect in such equations and thus always ignored, I don't think there is any geometrical or algebraic significance of this.
