In this post, the poster suspected that the $\log$ function would eventually flatten out and approach a straight line. We all know this isn't true of course. But then a commenter pointed out this:
@Ariel: yes, and yet in another sense it does tend to a flat line
because the slope approaches zero! Such is the way of limits. –
WChargin Aug 26 at 1:12
which hadn't occured to me. Is there any nice way of seeing how this is not a contradiction? I mean, I could imagine myself using the limit of derivatives to prove various stuff about the asymptotic behaviour of similar functions, for example boundedness. Why can't I?
Best Answer
The simplest way to see this is by taking $$\lim_{x \to \infty} \frac{d}{dx}\ln x = \lim_{x \to \infty} 1/x = 0$$ and as such observing that because the slope approaches zero $\ln x$ flattens out as $x \to \infty$. Unfortunately, this method offers zero intuition.
Similar behavior occurs in a discrete case with the harmonic series. A grouping technique such as this proves this series is divergent:
$$1 + 1/2 + 1/3 + 1/4 + \ldots $$
As for why derivatives cannot be used for bounding, imagine there is a continuous function (say, a smooth curve connecting each discrete point) which at $x = n, n \in \mathbb{N}$ increases by the nth term of the harmonic series from the last $x = n$ point. It is evident therefore that the slope (approximated by the difference between two discrete $x=n$ points and therefore approaching the nth term of the harmonic series) is constantly decreasing. Nonetheless, the integral still diverges, as even when every other point is disregarded except all $x \in \mathbb{N}$, we still have a divergent sum.
In this sense, the reason you can't prove boundedness for continuous functions is the same reason why you can't prove boundedness for the harmonic series by looking at the individual terms: even if the terms approach zero, they may not approach zero "fast enough" to be finite.
Boundedness cannot be proved using limits of derivatives, but can be proved using indefinite integrals. For $\ln x$, it goes as such:
$$\int_0^\infty\ln x \ dx = \lim_{a \to \infty}\big[x \cdot (\ln x - 1) \big]_0^a$$
This clearly diverges, as it is the product of two increasing terms.