Prove the remarkable limit of the function a^1/n, a>0, lim a^1/n=1 if n approaches to + infinite

To prove this remarkable limit, we'll prove that the limit
of ratio of 2 consecutive terms of the sequence (an) exists and it's value is
1.

lim a n+1/a n = lim
a^[1/(n+1)]/a^(1/n)

We'll use the quotient property of
exponentials that have matching bases:

lim
a^[1/(n+1)]/a^(1/n) =  lim a^[1/(n+1) - (1/n)]

lim
a^[1/(n+1) - (1/n)] = lim a^[-1/n(n+1)]

We'll use the
property of negative power:

lim a^[-1/n(n+1)] = lim
1/a^[1/n(n+1)]

lim 1/a^[1/n(n+1)] = 1/a^lim
[1/n(n+1)]

lim [1/n(n+1)] = 1/+infinite =
0

1/a^lim [1/n(n+1)] = 1/a^0 = 1/1 =
1

The limit of the ratio a n+1/ a n exists and it is
1.

Therefore, the remarkable limit, if n
approaches +infinite, is: lim a^(1/n) = 1.