The maximum value of $\frac{logx}{x}$ for x in [2, ∞) is :

$\frac{1}{e}$

The correct answer is Option (4) → $\frac{1}{e}$

$y=\frac{\log x}{x}$

so $\frac{dy}{dx}=\frac{1}{x^2}-\frac{\log x}{x^2}=0$

$⇒x=e$

$\frac{d^2y}{dx^2}=\frac{-2}{x^3}+\frac{2\log x}{x^3}-\frac{1}{x^3}$

$\frac{d^2y}{dx^2}=\frac{2}{e^3}-\frac{3}{e^3}<0$ 

so $x=e$ (maxima point)

$\left.\right]_{x=e}=\frac{1}{e}$ (max value)