The maximum value of (log x)/x for x in [2, ∞) is:

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

Let \(f(x) = \frac{\log x}{x}\). By the quotient rule,

    \(\displaystyle f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2}\)

    \(\displaystyle = \frac{1 - \log x}{x^2}\)

Setting f′(x) = 0:

    \(\log x = 1\)
    \(x = e\)

Since e ≈ 2.718, this point lies in [2, ∞).

For 2 ≤ x < e, log x < 1, so f′(x) > 0 (f increases); for x > e, log x > 1, so f′(x) < 0 (f decreases). So f has its greatest value at x = e:

    \(\displaystyle f(e) = \frac{\log e}{e} = \frac{1}{e}\)

Tip: e is the point where the maximum occurs, not the maximum value; e is Option (1).