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

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Answer 1 · 2015-09-11T00:17:03

It is equal to #1#

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 · 2015-09-11T00:38:03

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.

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#