Maximum value of $\left(\dfrac{1}{x}\right)^x$ is:

Let $\displaystyle y = \left(\dfrac{1}{x}\right)^x = e^{x\ln(1/x)} = e^{-x\ln x}$ Take $\ln$ on both sides: $\ln y = -x\ln x$ Differentiate w.r.t $x$: $\dfrac{1}{y}\dfrac{dy}{dx} = -(\ln x + 1)$ $\Rightarrow \dfrac{dy}{dx} = -y(\ln x + 1)$ For maximum or minimum, set $\dfrac{dy}{dx}=0$: $\ln x + 1 = 0 \Rightarrow x = \dfrac{1}{e}$ Now, $y_{\max} = \left(\dfrac{1}{1/e}\right)^{1/e} = e^{1/e}$