Domain of h2

The natural logarithm $\ln x$ requires $x > 0$. The denominator $x$ is already excluded from being $0$ by this same condition, so no further restriction is needed.

Therefore the domain is all positive real numbers, $x > 0$.

As $x \to \infty$, both $\ln x \to \infty$ and $x \to \infty$, so the limit has the indeterminate form $\infty/\infty$. This justifies applying L'Hopital's Rule.

Differentiating numerator and denominator separately: $\lim_{x\to\infty} \frac{\ln x}{x} = \lim_{x\to\infty} \frac{1/x}{1} = \lim_{x\to\infty} \frac{1}{x} = 0$.

Since this limit equals $0$, the graph has a horizontal asymptote $y = 0$ as $x \to \infty$.

Using the quotient rule, $h_2'(x) = \dfrac{(1/x)\cdot x - \ln x \cdot 1}{x^2} = \dfrac{1 - \ln x}{x^2}$.

Setting $h_2'(x) = 0$ gives $1 - \ln x = 0$, so $\ln x = 1$, i.e. $x = e$. For $0<x<e$, $h_2'(x) > 0$ (increasing), and for $x > e$, $h_2'(x) < 0$ (decreasing), confirming a local maximum at $x=e$.

The maximum value is $h_2(e) = \dfrac{\ln e}{e} = \dfrac{1}{e}$, and since this is the only critical point with the function increasing then decreasing, it is the global maximum on the domain.

As $x \to 0^+$, $\ln x \to -\infty$ while $x \to 0^+$ (a small positive number), so the numerator becomes very negative and the denominator becomes very small and positive.

This is not an indeterminate form since it's $(-\infty)/(0^+)$, which directly gives $\lim_{x\to 0^+} \frac{\ln x}{x} = -\infty$.

Since the function grows without bound (negatively) as $x$ approaches $0$ from the right, there is a vertical asymptote at $x = 0$.