How can one prove that $\lim_{n \to \infty}a^{1/n}=1$ for every $a>0$? Prove that $\lim_{n \to \infty} a^{\frac{1}{n}} = 1$ if $a >0$. 
In my textbook, we are given a suggestion to let $a^{\frac{1}{n}} = (1+h_n)$ and then show that the $h_n$ term goes to zero using a Theorem that states the following conditions:
$\lim_{n \to \infty} = \infty,$ if $a >1, 1,$ if $a = 1,$ and 0, if $\vert a \vert < 1$.
I apologize for not formatting the above cases appropriately; I could not figure out how to use a giant left brace to group them altogether.
If I rewrite it as $a^{n^{-1}}$, then I thought that this might help, but I don't think the Binomial Theorem would then apply since the exponent is negative. Furthermore, the problem is only concerned with the positive, real $n$-th roots. I am quite stuck on this problem and any suggestions or advice would be greatly appreciated. 
I am using the textbook Introduction to Analysis by Arthur Mattuck. 
 A: Case 1: Suppose that $a\gt 1$. Let $a=1+\delta$, where $\delta$ is positive.
We claim that if $n \ge 1$ then 
$$1\le a^{1/n}\lt 1+\frac{\delta}{n}.\tag{1}$$
The fact that $\lim_{n\to\infty} a^{1/n}=1$ is an immediate consequence of Inequality (1). 
To prove (1), suppose to the contrary that $a^{1/n}\gt 1+\frac{\delta}{n}$.  Then 
$$a=(a^{1/n})^n\gt \left(1+\frac{\delta}{n}\right)^n.\tag{2}$$
But by the Bernoulli Inequality,
$$\left(1+\frac{\delta}{n}\right)^n \ge 1+n\frac{\delta}{n}=1+\delta.\tag{3}$$
From (2) and (3) we conclude that $a\gt 1+\delta$, contradicting the fact that $a=1+\delta$.
Case 2: Suppose that $0\lt a\lt 1$. Let $b=\frac{1}{a}$. Then $b\gt 1$. By Case 1, $b^{1/n}$ has limit $1$, and therefore so does $a^{1/n}$.
Remarks: $1.$ The Bernoulli Inequality states that if $t\gt -1$, then $(1+t)^n\ge 1+nt$. We only need it for $t\gt 0$. In that case, it is an immediate consequence of the Binomial Theorem. There is also a quite  simple induction proof. 
$2.$ We gave a quite  formal proof. But it comes down to the fact that a number $\gt 1$ raised to a large enough power is very large, and in particular greater than $a$. 
A: Note:
I first saw this in "What is Mathematics?"
by Courant and Robbins.
You need two cases:
$0 < a < 1$
and $a > 1$.
These both use the motto
"Always expand around zero."
These cases both use
Bernoulli's inequality
(if $x \ge 0$ and $n \ge 1$
then $(1+x)^n \ge 1+xn$).
These also implicitly
use the Archimedean axiom
for the real numbers
(for any positive reals
$x$ and $y$
there is an integer $m$
such that
$mx > y$).
If $a > 1$,
let $a^{1/n} = 1+b$
where $b > 0$.
Then
$a 
=(1+b)^n
\ge 1+bn
$
so
$b 
\le \frac{a-1}{n}
$.
We can make $b < \epsilon$
for any $\epsilon$
by choosing 
$n
> \frac{a-1}{\epsilon}
$.
If $0 < a < 1$,
let $a^{1/n} = \frac1{1+b}$
where $b > 0$.
Then
$a = \frac1{(1+b)^n}$.
As before,
$(1+b)^n
\ge 1+bn
$,
so
$\begin{align}
&a 
\le
\frac1{1+bn}\\
&\text{or}\\
&\frac1{a}
\ge 1+bn\\
&\text{or}\\
&b 
\le \frac{1/a-1}{n}\\
&\text{or}\\
&1+b 
\le 1+\frac{1/a-1}{n}
= \frac{n+1/a-1}{n}\\
&\text{or}\\
&\frac1{1+b} 
\ge \frac{n}{n+1/a-1}
= 1-\frac{1/a-1}{n+1/a-1}\\
\end{align}
$
We can make
$\frac1{1+b} 
> 1-\epsilon
$
by making
$n+1/a-1
>\frac{1/a-1}{\epsilon}
$
or
$n
>(\frac1{a}-1)(\frac1{\epsilon}-1)
$
for any
$\epsilon > 0$.
A: Check this and then use the squeeze theorem with:
$$(1)\;\;a>1\implies 1\le\sqrt[n] a\le\sqrt[n]n\;\;,\;\;\text{for}\;\;n\ge a\\(2)\;\;a<1\implies\;\text{use arithmetic of limits with}\;\;b:=\frac1a>1\ldots$$
