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

2 Answers
Mar 29, 2017

Answer:

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#

Apr 2, 2017

Answer:

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.