Can you use the Big Theorem to compute limits of rational functions at infinity?

Question

The Big Theorem is used for sequences and natural numbers, but can we still apply it to continuous functions?

Let $a_n$ and $b_n$ be sequences. $a_n \ll b_n \implies \lim_{n\to\infty} \frac{a_n}{b_n} = 0$

The Big Theorem: $\ln n \ll n^a \ll c^n \ll n! \ll n^n, \forall a > 0, \forall c > 1$

It establishes a hierarchy among positive sequences.

Essentially, it's saying that the denominator approaches infinity faster than the numerator, even though both eventually approach infinity.

Now, let's say I wanted to compute this limit: $$\lim_{x\to \infty} \frac{x}{e^x}$$

Applying the Big Theorem to a similar problem, but with natural numbers, $$\lim_{n\to\infty}\frac{n}{e^n} = 0$$, for some natural number n.

I'm thinking that even though sequences only use natural numbers, they still approach the same number at infinity as its continuous counterpart. Is my assumption wrong? Are there any questions that would give the wrong answer if we used the Big Theorem instead of L'Hopital's law?

Note: There are some functions that can be defined such that the sequence of natural numbers of that function has a limit, but the continuous function doesn't have a limit (for example, let $f(x)$ be a sin function s.t. it has x-intercepts at 1, 2, 3, 4, ... . Let $a_n = f(n)$. Now, $\lim_{n \to \infty} a_n = 0$, but $\lim_{x \to \infty} f(x) = DNE$). I'll adjust my question 'not to include such examples', if that makes sense, because questions that ask to compute limits don't necessarily define such functions. My main question is with regards to simple functions, such as $\ln n, n^a, c^n, n!, n^n$.

Answer 1

We can extend in many cases this kind of results obtained for sequences also for continuous functions as follows using that $\forall x\in \mathbb R\;\exists n\in \mathbb Z$ such that $x\in [n,n+1)$. Notably, for the given example we have

$$ \frac{n}{e^{n+1}}\le \frac{x}{e^x} \le \frac{n+1}{e^n}$$

and then conclude by squeeze theorem.

As already noted in the comments, this is not possible in general (e.g. $a_n=\sin(2\pi n)$).

Answer 2

It seems to me that any proof of your "Big Theorem" could easily be adapted to prove a suitable continuous version, e.g. $$\forall a > 0, \forall c > 1,\ln x \ll x^a \ll c^x \ll \lfloor x\rfloor! \ll x^x$$ where $f(x)\ll g(x)$ means $\lim_{x\to+\infty}\dfrac{f(x)}{g(x)}=0$.