How to show that the limit $\lim_{n \in\mathbb N}\sqrt[n]n$ exists using epsilon-delta method. Finding the limit using L'Hopital's method might seem fine for me, but how can you prove that the limit exists : 

$\lim_{n \in\mathbb N}\sqrt[n]n$
It is clear that $(\forall n\in\mathbb{N}):\sqrt[n]n&gt;1$. Let, for each $n\in\mathbb N$, $\varepsilon_n=\sqrt[n]n-1$. Then, if $n&gt;1$,\begin{align*}n&amp;=\bigl(\sqrt[n]n\bigr)^n\\&amp;=(1+\varepsilon_n)^n\\&amp;=1+n\varepsilon_n+\frac{n(n-1)}2{\varepsilon_n}^2+\cdots\\&amp;&gt;\frac{n(n-1)}2{\varepsilon_n}^2.\end{align*}Therefore,$$(\forall n\in\mathbb{N}\setminus\{1\}):0&lt;\varepsilon_n&lt;\sqrt{\frac2{n-1}}$$and this proves that$$\lim_{n\in\mathbb N}\sqrt[n]n=\lim_{n\in\mathbb N}(1+\varepsilon_n)=1.$$
$$ \sqrt[n+1]{n+1}&lt; \sqrt[n]{n}$$

$$ (n+1)^{n} &lt; n^{n+1}$$

$$ n &gt; \left(\frac{n+1}{n}\right)^{n}= \left(1 +\frac{1}{n}\right)^{n},\ \  n &gt; n_{0}.$$

Sequence  $ \left(\left(1 + \frac{1}{n}\right)^{n}\right)$ is bounded.

Sequence $(\sqrt[n]{n})$ is decresing and lower bounded  number 1.

From Bolzano-Weierstrass theorem - sequence  $\sqrt[2n]{2n}$ is convergence to the same limit g.

$ g^2 = g\cdot g =\lim_{n\to \infty}\sqrt[2n]{2n}\cdot \lim_{n\to \infty}\sqrt[2n]{2n}= \lim_{n\to \infty}(\sqrt[2n]{2n})^{2}= \lim_{n\to \infty}(\sqrt[n]{2}\cdot \sqrt[n]{n}) = \lim_{n\to\infty}\sqrt[n]{2}\cdot \sqrt[n]{n}=1\cdot g $

$$ g = 1. $$