Now $x^{1/x}=e^{1/x\log x}$ and

${\displaystyle \frac{d}{\mathrm{dx}}(\frac{1}{x}\log x)=\frac{1-\log x}{x^2}.}$

The derivative is positive for $\log x<1$, that is $x<e$, it is is zero when $\log x=1$, $x=e$ and it is negative for $x>e$. Hence the function is increasing for $x<e$ and decreasing for $x>e$.

${\displaystyle \frac{d^2}{{\mathrm{dx}}^2}(\frac{1}{x}\log x)=\frac{-x-2x(1-\log x)}{x^4}=\frac{2\log x-3}{x^3}.}$

At $e$ this is $-1/e^3$ so $(e,e^{1/e})$ is a maximum point. As the graph is decreasing as $x\to \infty$ for each value of $x$ there is a value of $n$ for which $n^{1/n}<x^{1/x}$ and vice versa so both tend to the same limit which is 1.

Writing $t=1/x$,

${\displaystyle x^{1/x}={\left(\frac{1}{t}\right)}^t=\frac{1}{t^t}.}$

As $x\to 0$ we have $t\to \infty$ and $t^t\to \infty$ so

${\displaystyle x^{1/x}=\frac{1}{t^t}\to 0}$

.

Hence the graph of $y=x^{1/x}$ is always above the line $y=0$ although the function is undefined at $x=0$.