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

##### 2 Answers

See the solution process below:

#### Explanation:

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

The term

If we let

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

It is indeterminate.

#### Explanation:

Note that

We can try to use it as an algebraic object.

For example, some arithmetic operations are supported by the real projective line

#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

Taking a look at

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

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

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

The problem we have is that the

Basically, the expression