# What is root(oo)(125) ?

It is equal to $1$

#### Explanation:

Assuming that you mean sqrt(sqrt(sqrt(sqrt(...(sqrt125)

we can write as ${\left(125\right)}^{{\left(\frac{1}{2}\right)}^{n}}$ where $n$ is the number of square roots applied.

As $n \to + \infty$ then ${\left(\frac{1}{2}\right)}^{n} \to 0$ hence ${125}^{0} = 1$

Sep 11, 2015

I'm not sure what you intend to mean, but:

$\sqrt[\infty]{125} = {\lim}_{n \to \infty} \sqrt[n]{125} = 1$

or perhaps you mean the result of applying square root an infinite number of times, with the same result.

#### Explanation:

Inifiniteth root

The "infiniteth root" of $x$ could be defined as a limit:

$\sqrt[\infty]{x} = {\lim}_{n \to \infty} \sqrt[n]{x}$ where $n \in \mathbb{Z}$.

Then:

$\sqrt[\infty]{x} = \left\{\begin{matrix}1 & \text{if x > 0" \\ 0 & "if x = 0" \\ "undefined" & "if x < 0}\end{matrix}\right.$

So $\sqrt[\infty]{125} = 1$

Apply square root an infinite number of times

If we write ${\sqrt{}}^{\left(n\right)} \left(x\right)$ to mean $\sqrt{\sqrt{. . \left(\sqrt{x}\right) . .}}$, with $n$ $\sqrt{}$'s, then we could define:

${\sqrt{}}^{\left(\infty\right)} \left(x\right) = {\lim}_{n \to \infty} {\sqrt{}}^{\left(n\right)} \left(x\right)$

Again we find:

${\sqrt{}}^{\left(\infty\right)} \left(x\right) = \left\{\begin{matrix}1 & \text{if x > 0" \\ 0 & "if x = 0" \\ "undefined" & "if x < 0}\end{matrix}\right.$

So ${\sqrt{}}^{\left(\infty\right)} \left(125\right) = 1$