Find the maximum value of $\frac{\log x}{x},x>0$.

$\frac{1}{e}$

The correct answer is Option (3) → $\frac{1}{e}$

Let $f(x) =\frac{\log x}{x},x>0$. It is differentiable for all $x > 0$.

$f'(x)=\frac{x.\frac{1}{x}-\log x.1}{x^2}=\frac{1-\log x}{x^2}$ and 

$f''(x)=\frac{x^2(0-\frac{1}{x})-(1-\log x).2x}{x^4}=\frac{-3+2\log x}{x^3}$.

$f'(x)=0⇒\frac{1-\log x}{x^2}=0⇒1-\log x=0⇒\log x=1⇒x=e$

Therefore, $x = e$ is the only point where extremum may occur.

$f''(e)=\frac{-3+2\log e}{e^3}=\frac{-3+2×1}{e^3}=-\frac{1}{e^3}<0$

⇒ f has a local maxima at $x = e$ and local maxima value = $f(e) =\frac{\log e}{e}=\frac{1}{e}$.

Hence, the maximum value of $\frac{\log x}{x}$ is $\frac{1}{e}$.