Let me try to connect one of your questions with one of your other questions. As stated by Shalop, no such function exists. You can see this, for example as follows. Let $f(x)$ be a (positively) divergent function - a function such that $\lim\limits_{x\to\infty} f(x) = \infty$. Since you can define $g(x) := \log f(x)$, which diverges more slowly than $f(x)$, there is always a function which is "less divergent, but still divergent". Thus, there cannot exist a "least divergent" function.

This matter is exactly similar to the properties of an open set, where, for example, the interval $(0, \infty)$ has no smallest element. The derivative of a function measures its rate of change - a function whose derivative is very positive at a point $x$ is increasing very quickly at that point. Now, if we think of functions $f(x)$ like $x, \log x, \log(\log x), \ldots$, then we can look at how their derivatives behave.

In order for such a function to *stop* being divergent, intuitively, we need it to level out at some point. That is, its derivative - its rate of change - must become less than or equal to zero. Afterall, as soon as the derivative crosses zero, the function is no longer headed toward $+\infty$. In particular, the interval of values such that $f(x)$ no longer heads toward $+\infty$ has and end-point: $(-\infty, 0]$. Namely, the interval $(0, \infty)$ is the interval of values for which $f(x)$ will still be heading toward infinity - it's an open interval!

Granted, the description I gave is very non-rigorous, and only deals with increasing functions. To deal with functions like $x\exp(\sin(x))$ or $|x \sin(x)|$, we would need a different approach. Each still tends to $+\infty$ as $x \to \infty$, but their derivatives fluctuate between positive and negative values! To use the same approach, we could instead look at the largest value of these functions *so far*: the *envelope function*

$E(x) := \sup\limits_{0\leq y\leq x} f(x).$

Notice that $E(x)$ is now an increasing function: for all $x_2 \geq x_1 \geq 0$, $E(x_2)\geq E(x_1)$. Then in order for $f(x)$ not to be a divergent function, we require that $\lim\limits_{x\to\infty} E(x) < \infty$. But this can only happen if there is some point $x_0$ past which $E(x)$ stops increasing - *i.e.*

There exists $x_0 \geq 0$ such that there does not exist $x > x_0$ with $E(x) > E(x_0)$. If this is true, then there is an interval $I = (a,\infty)$ such that $E(x) \equiv 0$ on $I$. This means that $E(x)$ once again takes values in the closed interval $(-\infty, 0]$, so that $f(x)$ cannot tend to $+\infty$.