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.


1 Answer 1


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:

$$ \sup \{\sqrt[n]{n}:n \in \mathbb{N}\}=\sqrt[3]{3}$$


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