The maximum value of \(\frac{\log x}{x}\) in \((2, \infty)\) is

\[
\text { Let } y=\frac{\log x}{x}
\]
On differentiating w.r.t. x, we get
\[
\frac{d y}{d x}=\frac{x \cdot \frac{1}{x}-\log x \cdot 1}{x^{2}}=\frac{1-\log x}{x^{2}}
\]
For maxima, put \(\frac{d y}{d x}=0\)
\(\begin{array}{lc}\Rightarrow & \frac{1-\log x}{x^{2}}=0 \\ \Rightarrow & \log x=1 \\ \Rightarrow & x=e\end{array}\)
Now, \(\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x) 2 x}{\left(x^{2}\right)^{2}}\)
At \(\mathrm{x}=\mathrm{e}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx^{2 }} \leq 0, \text { maxima }}\)
\(\therefore\) The maximum value at \(\mathrm{x}=\mathrm{e}\) is
\[
\mathrm{y}=\frac{\log \mathrm{e}}{\mathrm{e}}=\frac{1}{\mathrm{e}}
\]