# What is root(oo)(oo) ?

Mar 29, 2017

See the solution process below:

#### Explanation:

We can rewrite this expression using this rule for roots and exponents:

$\sqrt[\textcolor{red}{n}]{x} = {x}^{\frac{1}{\textcolor{red}{n}}}$

$\sqrt[\textcolor{red}{\infty}]{\infty} = {\infty}^{\frac{1}{\textcolor{red}{00}}}$

The term $\frac{1}{\infty}$ approaches and equals $0$.

$\frac{1}{1} = 1$

$\frac{1}{2} = 0.5$

$\frac{1}{100} = 0.01$

$\frac{1}{1000000} = 0.000001$

$\frac{1}{10000000000000} = 0.0000000000001$

If we let $\frac{1}{\infty} = 0$ we can rewrite the expression as: ${\infty}^{0}$

We can then use this rule of exponents to complete the simplification of this expression:

${a}^{\textcolor{red}{0}} = 1$

${\infty}^{\textcolor{red}{0}} = 1$

Apr 2, 2017

It is indeterminate.

#### Explanation:

Note that $\infty$ is not really a number. It is more of a shorthand to express ideas like "as $n > 0$ gets larger without limit".

We can try to use it as an algebraic object.

For example, some arithmetic operations are supported by the real projective line ${\mathbb{R}}_{\infty} = \mathbb{R} \cup \left\{\infty\right\}$:

$\frac{1}{\infty} = 0$

$\frac{1}{0} = \infty$

$\infty + \infty = \infty$

If you do this, then you will find that there are cases which are indeterminate:

0 * oo = ?

oo - oo = ?

In calculus, instead of adding just one point to the real line $\mathbb{R}$, we effectively add two, namely $+ \infty$ (a.k.a $\infty$) and $- \infty$. Then we can speak of limits ${\lim}_{x \to \infty}$ or ${\lim}_{x \to - \infty}$.

Taking a look at $\sqrt[\infty]{\infty}$, the first question is "what does it mean?".

We can try to make sense of it with limits.

If we do then we find:

${\lim}_{m \to \infty} {\lim}_{n \to \infty} \sqrt[n]{m} = {\lim}_{m \to \infty} {\lim}_{n \to \infty} {m}^{\frac{1}{n}}$

$\textcolor{w h i t e}{{\lim}_{m \to \infty} {\lim}_{n \to \infty} \sqrt[n]{m}} = {\lim}_{m \to \infty} {m}^{0}$

$\textcolor{w h i t e}{{\lim}_{m \to \infty} {\lim}_{n \to \infty} \sqrt[n]{m}} = {\lim}_{m \to \infty} 1$

$\textcolor{w h i t e}{{\lim}_{m \to \infty} {\lim}_{n \to \infty} \sqrt[n]{m}} = 1$

We also find:

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

So $\sqrt[\infty]{\infty}$ looks like it should have the value $1$.

However, consider the the following definitions:

$\left\{\begin{matrix}{m}_{k} = {2}^{k} \\ {n}_{k} = k\end{matrix}\right.$

Then:

${\lim}_{k \to \infty} {m}_{k} = {\lim}_{k \to \infty} {n}_{k} = \infty$

${\lim}_{k \to \infty} \sqrt[{n}_{k}]{{m}_{k}} = {\lim}_{k \to \infty} {\left({2}^{k}\right)}^{\frac{1}{k}} = 2$

This is still some kind of $\sqrt[\infty]{\infty}$, so perhaps $2$ is a possible candidate value.

Consider the definitions:

$\left\{\begin{matrix}{m}_{k} = {2}^{{k}^{2}} \\ {n}_{k} = k\end{matrix}\right.$

Then:

${\lim}_{k \to \infty} {m}_{k} = {\lim}_{k \to \infty} {n}_{k} = \infty$

${\lim}_{k \to \infty} \sqrt[{n}_{k}]{{m}_{k}} = {\lim}_{k \to \infty} {\left({2}^{{k}^{2}}\right)}^{\frac{1}{k}} = {\lim}_{k \to \infty} {2}^{k} = \infty$

Oh dear! Looks like $\sqrt[\infty]{\infty} = \infty$ is also a possibility.

The problem we have is that the $\infty$'s represent two limit processes which are competing with one another. There is no obligation for the limit processes to track one another - they both just have to get larger and larger without limit.

Basically, the expression $\sqrt[\infty]{\infty}$ is indeterminate.