The maxmum value of $x^{1/x}$ is

$(1/e)^e$ $e^{1/e}$ $e$ $1/e$

$e^{1/e}$

Let $f(x) = x^{1/x} \log f(x) = (1/x) \log x$. Differentiating both the sides. we have $f’(x) = f(x)$. So $f’(x) = 0 x = e$. Also $f’(x) > 0$ for $0 < x < e$ and$ f’(x) < 0 for e < x < 0$. Thus f(x) has a maximum at x = e and max $f(x) = e^{1/e}$