Graph of the Function $y = \dfrac{\ln x}{x}$

$y = \dfrac{\ln x}{x}$ is a logarithm divided by a linear function. The numerator grows only slowly, so the denominator eventually overtakes it and the whole expression returns to $0$. The maximum reached along the way captures something essential about the number $e$.

Domain and range

The domain is $x > 0$. As $x \to 0^{+}$ the numerator tends to $-\infty$ and the denominator to $0^{+}$, so $y \to -\infty$; as $x \to \infty$ we get $y \to 0^{+}$. The maximum value is $\dfrac{1}{e}$, so the range is $y \leq \dfrac{1}{e}$.

Asymptotes and intercept

The $y$-axis is a vertical asymptote and the $x$-axis a horizontal one, though on the right the curve approaches it from above. Since $\ln x = 0$ at $x = 1$, the $x$-intercept is $(1, 0)$; the function is negative on $0 < x < 1$ and positive for $x > 1$.

Monotonicity and extrema

The quotient rule gives the derivative.

The denominator is positive, so the sign comes from $1 - \ln x$: positive for $x < e$ and negative for $x > e$. The function therefore has a maximum, both local and global, at $x = e$, of value $\dfrac{1}{e}$. The maximum point is $\left( e, \dfrac{1}{e} \right) \approx (2.718, 0.368)$.

Concavity and inflection

The second derivative is $y'' = \dfrac{2\ln x - 3}{x^3}$. The denominator is positive, so the sign is that of $2\ln x - 3$, which changes at $x = e^{3/2} \approx 4.482$. That single inflection point, where the curve turns from concave down to concave up, has value $\dfrac{3}{2}e^{-3/2} \approx 0.335$.

Comparing $e^{\pi}$ and $\pi^{e}$

Because the maximum sits at $x = e$, a famous comparison follows. For $x \neq e$ we have $\dfrac{\ln x}{x} < \dfrac{1}{e}$, and taking $x = \pi$ gives the following chain.

Numerically $\pi^{e} \approx 22.459$ and $e^{\pi} \approx 23.141$. The same argument shows that $a^{b} > b^{a}$ whenever $e \leq a < b$.

Relation to $\sqrt[n]{n}$

Since $x^{1/x} = e^{(\ln x)/x}$, the maximum point of this function is also the maximum point of $x^{1/x}$, which is $e^{1/e} \approx 1.445$ at $x = e$.

Among the integers the closest to $e$ is $3$, so the sequence $\sqrt[n]{n}$ peaks at $n = 3$. As the graph settles toward $0$, the sequence tends to $1$.

Integral

Substituting $t = \ln x$ gives a tidy result.