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

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

We use indeterminate forms because it helps us identify particular groups of problems that we can tackle in a singular way. It's a pretty common tool in maths.

Think about derivatives. Maybe you first learned about finding the derivative via "first principles", i.e. evaluating

$\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$

So you do that for some basic polynomials, a couple of trigonometric functions, maybe the exponential or logarithm functions, and you get a feel for how it works. But if you were asked to differentiate $f(x) = \frac{xe^x}{\sin(x^2 - 4)}$, are you going to write

$\lim_{h \rightarrow 0} \frac{\frac{(x+h)e^{x+h}}{\sin((x+h)^2 - 4)} - \frac{xe^x}{\sin(x^2 - 4)}}{h}$

then rearrange it and try to make it something usable? I sincerely hope not. Instead, you'll go "Well there's a fraction of two functions, so I'll use the Quotient Rule, then in the numerator there's two functions multiplied together, so I'll use the Product Rule on that, then in the denominator there's a function of a function so I'll use the Chain Rule". By identifying certain "classes" of functions, you can define rules to apply to find their derivatives.

It's the same with limits. You already know that if you have the quotient of two functions, $\frac{f(x)}{g(x)}$, and you take a limit where they both go to a finite value (and $g(x)$ doesn't go to zero), then you can just substitute those values. Why? Because there's a theorem that if $f(x) \rightarrow a$ and $g(x) \rightarrow b \neq 0$, then $\frac{f(x)}{g(x)} \rightarrow \frac{a}{b}$.

Similarly, if only one of the two functions follows that rule, then there's another theorem you can apply to say where the limit goes. For example, if in a particular limit $f(x) \rightarrow \infty$ and $g(x) \rightarrow b$ (where, remember, a function "approaching infinity" is just shorthand for "grows without bound and hence has no finite limit"), then $\frac{f(x)}{g(x)} \rightarrow \infty$.

Indeterminate forms happen to be a class of limits where you can't apply a single rule, because for any kind of indeterminate form you can find a particular instance that has any limit you choose. For example, in the case of $\frac{\infty}{\infty}$, we have $\lim_{x \rightarrow \infty}\frac{x}{kx} = \frac{1}{k}$, $\lim_{x \rightarrow \infty}\frac{x^2}{kx} = \infty$ and $\lim_{x \rightarrow \infty}\frac{x}{kx^2} = 0$.

So maybe you define a rule if you're working with polynomials, but then what if someone throws a logarithm or exponential function into the mix? Or a trig function? Sure you can say "$\log(x)$ grows slower than $x$", but how do you know? How does $xe^{x^2}$ go against $x^2 e^x$? Does $\sin(e^x) \log(x)$ even do anything sensible?

The nice thing is that we can tackle an extremely broad range of these indeterminate forms with a single rule[1]. So the fact that we can (1) define a class of limits and (2) determine the behaviour of all members of that class with one method is why we use indeterminate forms. If we didn't have l'Hopital's rule we probably wouldn't mention them so much.

[1] Actually it's a bunch of related rules and they only apply if the functions involved are sufficiently well-defined, but that still happens to be good enough for a wide variety of functions that we actually care about.