1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You might find it easier to consider the function $x\mapsto \sqrt x$ instead. This also flattens out, and the slope approaches $0$. But I think it's more intuitively obvious that $\sqrt x$ is not bounded. $\endgroup$ – davidlowryduda Sep 1 '15 at 6:46
  • $\begingroup$ It is intuitively obvious that the logarithm is unbounded as well. That's not the question. $\endgroup$ – Benjamin Lindqvist Sep 1 '15 at 6:49
  • 1
    $\begingroup$ I'm a bit taken aback by the curtness of your reply, and I do not plan on responding further. But this is purely a problem of intuition. You ask why you cannot do something which you know to be impossible. Alternately, you ask why two easily proved statements do not contradict each other. You have a false impression that if a function becomes flatter, it should be bounded. I do not know why you think this. Perhaps you should think of this as sharpening your intuition. You might benefit from meditating on further counterexamples in analysis, too. $\endgroup$ – davidlowryduda Sep 1 '15 at 6:55
  • $\begingroup$ Counterexamples are not always the most illuminating ways to think about concepts. What if we run into a different scenario where we want to use the asymptotic behaviour of derivatives to prove stuff other than boundedness about a function? What if we don't have a counterexample in that case? How will we know if we are allowed to utilize asymptotics of derivatives? $\endgroup$ – Benjamin Lindqvist Sep 1 '15 at 7:03
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This answer also misses the gist of the question. The question is "what is the logical flaw in using limits of derivatives to prove boundedness?" $\endgroup$ – Benjamin Lindqvist Sep 1 '15 at 6:53
  • $\begingroup$ Ahh. Gotcha. I'll update my answer. $\endgroup$ – Chandler Watson Sep 1 '15 at 6:57
  • $\begingroup$ This is a very nice edit! $\endgroup$ – Benjamin Lindqvist Sep 1 '15 at 7:42
0
$\begingroup$

"Approaches a straight line" here is only a figure of speech.There is no straight line f(x)=Ax+B (with constant A,B) such that |f(x)-log(x)| tends to zero as x goes to infinity.(A hyperbola has asymptotic lines. The log doesn't.) However since the log is monotonic and its derivative does converge to zero,this means that if we take x large enough, the graph of log (y) for, say, x-1 < y < x+1 will be very close to the line joining (x-1,log(x-1) to (x+1,log (x+1). But if we replace x with a much greater value, we will need a different line to get a good "local fit" to the graph of the log. There is no contradiction,when you specify exactly what you mean by "approaches a line."

$\endgroup$
0
$\begingroup$

What is true of the derivative is not always true of the main integrand function.

Although $\frac{1}{x} \rightarrow 0,$ log(x) does not tend to some limit asymptotically, we have to test each case about boundedness, graphical appearances could be misleading.

$\endgroup$
0
$\begingroup$

Of course, it is wellknown that a necessary condition to a function $f(x)$ tends to a finite limit when $x$ tends to infinity is that $f'(x)$ tends to $0$. This is a necessary condition, but not a sufficient condition.

The opposite statement is not true : If $f'(x)$ tends to $0$ when $x$ tends to infinity, this doessn't imply that $f(x)$ tends to a finite limit. They are a lot of examples, such as $f(x)=x^a$ with $0<a<1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.