Supremum of $\{\sqrt[n]{n}:$ $n \in \mathbb{N}\}$ and $\{\sqrt[x]{x}:$ $x \in \mathbb{R}_{>0}\}$

I would like to calculate the supremums of these sets:

$\{\sqrt[n]{n}:$ $n \in \mathbb{N}\}$ and $\{\sqrt[x]{x}:$ $x \in \mathbb{R}_{>0}\}$

I don't know how to start, any help is appreciated.

For $x^\frac{1}{x}$, if you take derivative with respect to $x$
$$\frac{\partial}{\partial x}\sqrt[x]{x}= x^{\frac{1}{x}-2}(-\ln x+1)$$
, and set to zero you get $\ln~x=1, \Rightarrow x=e$. Therefore:
$$\sup \{\sqrt[x]{x}:x \in \mathbb{R}_{>0}\}=\sqrt[e]{e}$$
Note that for $x<e$ the deivative is positive (the function is ascending), and for $x>e$ the function is descending ($x=e$ is the maximum); therefore, for integer values, the suprememum is either on $x=2$ or $x=3$, by checking we get $2^{\frac{1}{2}}=1.4142$, and $3^{\frac{1}{3}}=1.4422$, therefore:
$$\sup \{\sqrt[n]{n}:n \in \mathbb{N}\}=\sqrt[3]{3}$$