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I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L < \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?

It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:

$$ \begin{align} \lim_{x \to +\infty} f'(x) &= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\ &= \lim_{h \to 0} \frac1{h}(L-L) \\ &= \lim_{h \to 0} \frac{0}{h} \\ &= 0 \end{align} $$

In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is a similar question already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case.

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  • $\begingroup$ @StubbornAtom: $\lim_{x\to\infty} f'(x) = 0$; it says so right there in the question. $\endgroup$
    – Javier
    Jun 11, 2016 at 15:36
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    $\begingroup$ If $\lim_{x\to\infty}f'(x)$ exists, then existence of $\lim_{x\to\infty}f(x)$ implies $f'\to 0$. To see this, apply the Mean Value theorem to show that there exists a $\xi \in (x,x+1)$ such that $f(x+1)-f(x)=f'(\xi)$. Now, let $x\to \infty$. $\endgroup$
    – Mark Viola
    Jul 8, 2021 at 19:33

6 Answers 6

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The answer is: No. Consider $f(x)=x^{-1}\sin(x^3)$ on $x\gt0$. The derivative $f'(x)$ oscillates between roughly $+3x$ and $-3x$ hence $\liminf\limits_{x\to+\infty}\,f'(x)=-\infty$ and $\limsup\limits_{x\to+\infty}\,f'(x)=+\infty$.

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    $\begingroup$ This is the answer I like more, simply because you provided a function for which it is easy to check that it's a counterexample (just differentiate and take limits). Thanks! $\endgroup$
    – Javier
    Jun 23, 2012 at 19:39
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    $\begingroup$ Also, a nitpick: shouldn't it be $x > 0$ instead of $x \ge 0$? $\endgroup$
    – Javier
    Jun 23, 2012 at 19:39
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    $\begingroup$ @Javier Thanks. Including $x=0$ is not a problem but since this could distract the reader, I modified the answer. $\endgroup$
    – Did
    Nov 16, 2015 at 7:30
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    $\begingroup$ What do you mean it's not a problem you are dividing by x $\endgroup$
    – Prince M
    Feb 14, 2017 at 3:19
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    $\begingroup$ I find not dividing by 0 to be useful $\endgroup$
    – Prince M
    Feb 14, 2017 at 9:46
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There is a famous theorem known as Barbalat's lemma, which states the additional condition for $\lim_{x \to \infty} f'(x) = 0$. According to the lemma, $f'(x)$ should be uniformly continuous on $[a, \infty)$. In many applications, the uniform continuity of $f'(x)$ is shown by proving $f''(x)$ exists and is bounded on $[a, \infty)$.

(See Wikipedia https://en.wikipedia.org/wiki/Lyapunov_stability#Barbalat.27s_lemma_and_stability_of_time-varying_systems for the statement of Barbalat's lemma and its applications in stability analysis).

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    $\begingroup$ This is a good point, but should really be a comment, not an answer. $\endgroup$ Nov 16, 2015 at 7:55
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    $\begingroup$ This can be definitely the general answer. All of the derivatives of the functions in the above counter examples are not uniformly continuous. As far as I know, there is no more general theorem than Barbalat's lemma. The lemma does not assume that the limit $\lim_{t \to \infty} f'(t)$ exists as well. $\endgroup$ Nov 17, 2015 at 16:46
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    $\begingroup$ This answer addresses the spirit of the question although it is not the answer in the strict sense. $\endgroup$
    – deps_stats
    Dec 6, 2019 at 22:50
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    $\begingroup$ In fact, if $\lim_{x\to\infty}f'(x)$ exists, then the existence of $\lim_{x\to\infty}f(x)$ implies $f'\to 0$. To see this, apply the Mean Value theorem to show that there exists a $\xi \in (x,x+1)$ such that $f(x+1)-f(x)=f'(\xi)$. Now, let $x\to \infty$. $\endgroup$
    – Mark Viola
    Jul 8, 2021 at 19:29
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Take a function that is $0$ except in a small neighborhood of each positive integer; at $n\in\Bbb Z^+$ it has a smooth bump of height and width $1/n$ whose rising part has a maximum slope of $n$. This function is differentiable and has limit $0$ at infinity, but its derivative has no limit at infinity.

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    $\begingroup$ What is "nbhd"? $\endgroup$
    – Hammerite
    Jun 23, 2012 at 19:54
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    $\begingroup$ "Neighborhood". $\endgroup$
    – Mat
    Jun 23, 2012 at 19:56
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Because all counterxamples given here are oscillating, perhaps someone might wonder whether the proposition is true if we require the funcion to be monotonic. The answer is still no. Simply consider the inverse of the function, monotonic on $(0,\infty)$, given here $f(x)=\frac{1}{x}+\sin\left(\frac{1}{x}\right)$.

Also: consider the cumulative distribution function corresponding to a probability density function that has no limit as $x\to +\infty$. For example, the integral of the density constructed here. Or, using the same idea with a Cauchy distribution we get other example:

$$F(x)= \frac{2}{\pi}\sum_{k=1}^\infty \frac{\tan^{-1}\left((x-k) \pi 2^k\right)}{2^k} $$

enter image description here

This tends to $1$, but its derivative $f(x)=F'(x)= 2 \sum_{k=1}^\infty (\pi^2 2^{2k} (x-k)^2 +1)^{-1} $ has no limit, because $f(n)>2$ $\forall n \in \mathbb{N}$.

enter image description here

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    $\begingroup$ Your first counterexample is invalid, since $\lim\limits_{x \to +\infty}f'(x)=\lim\limits_{x \to +\infty}-\frac{1}{x^2}\left(\cos\frac{1}{x}+1\right)=0.$ $\endgroup$ Nov 16, 2019 at 20:38
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    $\begingroup$ @mengdie1982 I don't think so. See the graph here math.stackexchange.com/questions/1863341/… $\endgroup$
    – leonbloy
    Nov 16, 2019 at 21:23
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    $\begingroup$ Sir,this is just a product of an infinitesimal and a bounded variable,so it tends to zero. Don't depend on the intuition too much. $\endgroup$ Nov 17, 2019 at 7:38
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    $\begingroup$ @mengdie1982 You are right in that (I misread what you wrote), but the example does not refer to $f(x)$ but to its inverse. $\endgroup$
    – leonbloy
    Nov 18, 2019 at 1:50
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Let a function oscillate between $y=1/x$ and $y=-1/x$ in such a way that it's slope oscillates between $1$ and $-1$. Draw the picture. It's easy to see that such functions exist. Then the function approaches $0$ but the slope doesn't approach anything.

One could ask: If the derivative also has a limit, must it be $0$? And there, I think, the answer is "yes".

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    $\begingroup$ Agree with the last line. If the limit of the derivative is (wlog) $a>0$, then from some point on, $f'(x)>a/2$ always. Together with the assumption that $f$ has a limit, this easily leads to a violation of the mean value theorem. $\endgroup$ Jun 23, 2012 at 17:51
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Not true. Here's how to construct a counterexample:

Let $f$ be a smooth function which satisfies the following:

  1. $f$ is zero everywhere except on intervals of the form $$ \left(n-\frac{1}{2n},n+\frac{1}{2n}\right), n \in \mathbb{N} $$
  2. On those intervals, smoothly rise from zero until $f(n)=\frac{1}{n}$, then fall back to zero.
  3. You can find the average slope on the interval $(n-\frac{1}{2n},n)$ is 2, and so by the Mean Value Theorem, this slope will be achieved on that interval.

Therefore, $f'(x)$ will not have limit zero since it will reach as high as 2 near each positive integer.

Addendum: interchanging limits is usually quite dangerous business. The most important theorems in real analysis involve, at their core, special situations under which we are allowed to interchange limits.

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