Why is $\frac{\ln\infty}{\infty}$ equal to $\frac\infty\infty$?

$\dfrac{\infty}{\infty}$ is a symbol that in this context just means the numerator and denominator both "approach infinity" (grow without bound). It's true that the numerator approaches infinity far slower than the denominator, which is why ${\displaystyle \lim_{x\to\infty}}\dfrac{\ln x}{x}=0$.

$\dfrac{\infty}{\infty}$ is called an "indeterminate form" because it doesn't tell you enough information to determine what the limit is.

Notice that for $x \ge 1$

$$0 \le \frac {\ln x} x \le \frac {\sqrt x} x = \frac 1 {\sqrt x}$$

and the last fraction tends to $0$ when $x \to \infty$, which means that $\frac {\ln x} x \to 0$ too. This shows that even though both $\ln x$ and $x$ tend to $\infty$, your intuition that $\ln x$ is much slower than $x$ is correct. Nevertheless, "evaluating" the fraction $\frac {\ln x} x$ directly in $\infty$ leads to the undefined object $\frac {\ln \infty} \infty = \frac \infty \infty$, therefore in analysis we never "evaluate at infinity", but rather "take the limit at infinity".

This is a simple case of an indeterminate form, more specifically, $\frac{\infty}{\infty}$ Using L'Hopital's's rule, differentiate the numerator and denominator

$$\lim_{x\to \infty}\frac{\ln x}{x} = \lim_{x\to \infty}\frac{\frac{d}{dx}\ln x}{\frac{d}{dx}x} = \lim_{x\to \infty} \frac{\frac{1}{x}}{1}=\lim_{x\to \infty}\frac{1}{x}=0$$