Use $\delta-\epsilon$ to show that $\lim_{n\to\infty} a^{\frac{1}{n}} = 1$?

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

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

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 $a>0$.