Use $\delta-\epsilon$ to show that $\lim_{n\to\infty} a^{\frac{1}{n}} = 1$? Hope this is a meaningful question, but I'm curious if is possible to show that 
$$\lim_{n\to\infty} a^{\frac{1}{n}}=1, \text{where  }a>0$$
using $\delta-\epsilon$ directly or other methods. One method that I am aware of is to use the following:

If $\{s_n\}$ is a nonzero sequence, then $\liminf\bigl|\frac{s_{n+1}}{s_n}\bigr|\le \liminf |s_n|^{\frac{1}{n}}\le \limsup |s_n|^{\frac{1}{n}}\le\limsup\bigl|\frac{s_{n+1}}{s_n}\bigr|$

 A: The case $a=1$ is obvious.Let now $a>1$,then $\sqrt [ n ]{ a } >1$ and 

$$a={ \left( 1+\left( \sqrt [ n ]{ a } -1 \right)  \right)  }^{ n }>1+n\left( \sqrt [ n ]{ a } -1 \right) >n\left( \sqrt [ n ]{ a } -1 \right) $$ (Bernouli's inequality was used )

from here we get that $0<\sqrt [ n ]{ a } -1<\frac { a }{ n } <\varepsilon $  when $n>\frac { a }{ \varepsilon  } ,\left( \varepsilon >0 \right) $ so $\sqrt [ n ]{ a } \rightarrow 1,n\rightarrow \infty $ 
now let consider that $0<a<1$ we have $\frac { 1 }{ a } >1$ and in this case we have also $\sqrt [ n ]{ \frac { 1 }{ a }  } \rightarrow 1,n\rightarrow \infty $   so that 

$$\lim _{ n\rightarrow \infty  }{ \sqrt [ n ]{ a } = } \lim _{ n\rightarrow \infty  }{ \frac { 1 }{ \sqrt [ n ]{ a }  } =\frac { 1 }{ \lim _{ n\rightarrow \infty  }{ \sqrt [ n ]{ a }  }  } =1 } $$

A: Case I: For $a=1$, the sequence converges to $1$.
Case II: If $a>1$, then $a^{1/n}>1$. Let $a^{1/n}=1+x_n$ where $x_n>0$.
Then $a=(1+x_n)^n=1+nx_n+...+x_n^n>nx_n \quad \forall  n\in \mathbb{N}$
$\therefore 0<x_n<\frac{a}{n} \quad \forall n\in \mathbb{N}$
As $\lim \frac{a}{n}=0$, it follows from the Sandwich theorem that $\lim x_n=0$.
Hence $\lim a^{1/n}=1$
Case III: For $0<a<1$, let $b=\frac{1}{a}$. Then $b>1$ and $\lim a^{1/n}=\lim\frac{1}{b^{1/n}}=1$ (from the 2nd case).
Combining all three cases, we have $\lim a^{1/n}=1$, if $a>0$.
A: Hint: You might want to use that $$a^{\log_a x}=x$$ and note that as  $\log_ax\rightarrow 0$,  $x\rightarrow 1$.
Yours is just a formalization of this, applied to $x=1-\epsilon$ if $a<1$, and $x=1+\epsilon$, if $\alpha>1$, using the fact that $\log$ is monotone. ($\alpha=1$ is clear)
EDIT: if you need more hints, or a complete solution, let me know.
