Question

What is `root(oo)(125)` ?

Answer 1

It is equal to `1`

Explanation:

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

we can write as `(125)^((1/2)^n)` where `n` is the number of square roots applied.

As `n->+oo` then `(1/2)^n->0` hence `125^(0)=1`

Answer 2

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

`root(oo)(125) = lim_(n->oo)root(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:

`root(oo)(x) = lim_(n->oo) root(n)(x)` where `n in ZZ`.

Then:

`root(oo)(x) = { (1, "if x > 0"), (0, "if x = 0"), ("undefined", "if x < 0") :}`

So `root(oo)(125) = 1`

Apply square root an infinite number of times

If we write `sqrt()^((n))(x)` to mean `sqrt(sqrt(..(sqrt(x))..))`, with `n` `sqrt()`'s, then we could define:

`sqrt()^((oo))(x) = lim_(n->oo) sqrt()^((n))(x)`

Again we find:

`sqrt()^((oo))(x) = { (1, "if x > 0"), (0, "if x = 0"), ("undefined", "if x < 0") :}`

So `sqrt()^((oo))(125) = 1`