So let's say I have some function $f(x)$ where both $f(x)$ and $x$ are constrained to be real positive integers, $1, 2, 3, \ldots$.
I was trying to find potential limits of this function. I realized that given those conditions, $f(x)$ will always be greater than $\ln(f(x))$.
So, I set: $f(x) > \ln(f(x))$
And, if I solve this for $f(x)$, I get the following steps
$$f(x) > \ln(f(x))$$
$$e^{f(x)} > f(x)$$
$$1 > f(x)e^{-f(x)}$$
$$-1 < -f(x)e^{-f(x)}$$
$$W(-1) < -f(x)$$
$$-W(-1) > f(x)$$
And because $-W(-1)$ appears to be an upper limit, and with the integer constraint then
$$\lfloor -W(-1) \rfloor \geq f(x)$$
But $W(-1)$ is a complex number... how am I supposed to understand what this is saying?
You have $-1<-f(x)e^{-f(x)}$.
The problem is when you say $W(-1)<W\left(-f(x)e^{-f(x)}\right)=-f(x)$.
This is only possible if $W(x)$ is increasing, real-valued and defined at both values for the inequality.
For example, if we have $-1<0$, we can't say $\sqrt{-1}<\sqrt 0\equiv i<0$ (not real-valued).
Another example is for $f(x)=-\frac{1}{x+1}$, we can't say $-1<0\implies f(-1)<f(0)\equiv -\frac{1}{0}<-1$ (not defined).
Another example is $f(x)=-x$, we can't say $-1<0\implies f(-1)<f(0)\equiv 1<0$, but we should say $f(-1)>f(0)\equiv 1>0$ because $f(x)$ is decreasing not increasing.
The problem is not when you apply $W$ to $-f(x)e^{-f(x)}$, but when you apply it to $-1$ because it's not real-valued there.