TITLE: Use $\delta-\epsilon$ to show that $\lim_{n\to\infty} a^{\frac{1}{n}} = 1$?
QUESTION [5 upvotes]: 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 $\liminf\bigl|\frac{s_{n+1}}{s_n}\bigr|\le \liminf |s_n|^{\frac{1}{n}}\le \limsup |s_n|^{\frac{1}{n}}\le\limsup\bigl|\frac{s_{n+1}}{s_n}\bigr|$

REPLY [6 votes]: 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 

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