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Show that $lim_limits{n rightarrow infty}sqrt[n]{n}=1$ using Bernoulli inequality

I have a hint $epsilon > 0$ and $n=ncdot epsilon/epsilon$ and I need to show $$1leq sqrt[n]{n}leq frac{1}{sqrt[n]{epsilon}}(1+epsilon)$$

What I have done so far is verifying the inequality on the left.
Also $$sqrt[n]{n}= frac{1}{sqrt[n]{epsilon}}sqrt[n]{epsilon}sqrt[n]{n}$$

Thanks for taking your time

$(1+frac1{sqrt{n}})^n
ge 1+frac{n}{sqrt{n}}
gt n^{1/2}
$.

Raising to the $2/n$ power,
$n^{1/n}
=(n^{1/2})^{2/n}
lt ((1+frac1{sqrt{n}})^n)^{2/n}
= (1+frac1{sqrt{n}})^2
=1+frac{2}{sqrt{n}}+frac1{n}
lt 1+frac{3}{sqrt{n}}
to 1
$.

This is how you are taking advantage of Bernoulli in order to obtain the second inequality:

$frac{(1+epsilon)^n}{epsilon}overset{Bernoulli}{geq} frac{1+nepsilon}{epsilon} = frac{1}{epsilon}+n geq n Rightarrowfrac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq n^{frac{1}{n}}$

Note that: $$lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}}=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}left (frac{1}{epsilon^{frac{1}{n}}}+frac{epsilon}{epsilon^{frac{1}{n}}}right ) right )=lim_{epsilonrightarrow 0}left (lim_{nrightarrowinfty}frac{1}{epsilon^{frac{1}{n}}}+lim_{nrightarrowinfty}frac{epsilon}{epsilon^{frac{1}{n}}} right)$$ $$=lim_{epsilonrightarrow 0}(lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}+lim_{nrightarrowinfty}epsilon^{1-frac{1}{n}})=lim_{epsilonrightarrow0}(1+epsilon)=1$$
$lim_{nrightarrowinfty}epsilon^{-frac{1}{n}}=1$ because of $epsilon$ being a constant.

To conclude:
$1=lim_{epsilonrightarrow0}lim_{nrightarrowinfty}frac{(1+epsilon)}{epsilon^{frac{1}{n}}} geq lim_{epsilonrightarrow0}lim_{nrightarrowinfty}n^{frac{1}{n}}=lim_{nrightarrowinfty}n^{frac{1}{n}}$

You can prove it like that: $(sqrt n)^{1/n}=1+k_n$
Where $k_n$ is some positive number depending on $n$

Then $sqrt n =(1+k_n)^n geq nk_n$-Bernoulli

So $k_n leq sqrt n/n =1/sqrt n$

And then $1 leq n^{1/n} =(1+k_n)^2=1+2k_n+k_n^2<1+2/sqrt n + 1/n$

And you get what you need using squeeze theorem

Option.

$x ge 0.$

$(star)$ $(1+x)^n gt (n^2/4)x^2$.

$x=2/√n.$

$(1+2/√n)^n gt n$

$1+2/√n gt n^{1/n} gt 1.$

Proof of $(star).$

$(1+x)^n ge n(n-1)x^2/2...gt$

$(n^2/4)x^2,$ since for $n >2$:

$(n-1) > n/2$.

As shown in this answer, using Bernoulli's Inequality, $left(1+frac1nright)^{n+1}$ is decreasing.

Thus, for $nge1$,
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac1{n+1}left(1+frac1nright)^{n+1}\
&lefrac4{n+1}
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&=left(frac{(n+1)^n}{n^{n+1}}right)^{frac1{n(n+1)}}\
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\[12pt]
&le1
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is decreasing and bounded below by $1$. Therefore, $limlimits_{ntoinfty}n^{1/n}$ exists and is no less than $1$.

Let $alpha=limlimits_{ntoinfty}n^{1/n}$, then
$$
begin{align}
alpha
&=lim_{ntoinfty}n^{frac1n}\
&=lim_{ntoinfty}(2n)^{frac1{2n}}\
&=lim_{ntoinfty}2^{frac1{2n}}lim_{ntoinfty}n^{frac1{2n}}\[3pt]
&=1cdotsqrt{alpha}
end{align}
$$
Since $alphage1$ and $alpha=sqrt{alpha}$, we get
$$
begin{align}
lim_{ntoinfty}n^{frac1n}
&=alpha\
&=1
end{align}
$$