Question

What is `root(oo)(oo)` ?

Answer 1

See the solution process below:

Explanation:

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

`root(color(red)(n))(x) = x^(1/color(red)(n))`

`root(color(red)(oo))(oo) = oo^(1/color(red)(00))`

The term `1/oo` approaches and equals `0`.

`1/1 = 1`

`1/2 = 0.5`

`1/100 = 0.01`

`1/1000000 = 0.000001`

`1/10000000000000 = 0.0000000000001`

If we let `1/oo = 0` we can rewrite the expression as: `oo^0`

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

`a^color(red)(0) = 1`

`oo^color(red)(0) = 1`

Answer 2

It is indeterminate.

Explanation:

Note that `oo` 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 `RR_oo = RR uu { oo }`:

`1/oo = 0`

`1/0 = oo`

`oo + oo = oo`

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 `RR`, we effectively add two, namely `+oo` (a.k.a `oo`) and `-oo`. Then we can speak of limits `lim_(x->oo)` or `lim_(x->-oo)`.

Taking a look at `root(oo)(oo)`, 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->oo) lim_(n->oo) root(n)(m) = lim_(m->oo) lim_(n->oo) m^(1/n)`

`color(white)(lim_(m->oo) lim_(n->oo) root(n)(m)) = lim_(m->oo) m^0`

`color(white)(lim_(m->oo) lim_(n->oo) root(n)(m)) = lim_(m->oo) 1`

`color(white)(lim_(m->oo) lim_(n->oo) root(n)(m)) = 1`

We also find:

`lim_(n->oo) root(n)(n) = 1`

So `root(oo)(oo)` looks like it should have the value `1`.

However, consider the the following definitions:

`{ (m_k = 2^k), (n_k = k) :}`

Then:

`lim_(k->oo) m_k = lim_(k->oo) n_k = oo`

`lim_(k->oo) root(n_k)(m_k) = lim_(k->oo) (2^k)^(1/k) = 2`

This is still some kind of `root(oo)(oo)`, so perhaps `2` is a possible candidate value.

Consider the definitions:

`{ (m_k = 2^(k^2)), (n_k = k) :}`

Then:

`lim_(k->oo) m_k = lim_(k->oo) n_k = oo`

`lim_(k->oo) root(n_k)(m_k) = lim_(k->oo) (2^(k^2))^(1/k) = lim_(k->oo) 2^k = oo`

Oh dear! Looks like `root(oo)(oo) = oo` is also a possibility.

The problem we have is that the `oo`'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 `root(oo)(oo)` is indeterminate.