# If a function has a finite limit at infinity, does that imply its derivative goes to zero?

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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Did you mean $f'(x)=0$ or $\lim_{x\to \infty}f'(x)=0$? – StubbornAtom Jun 11 at 11:04
@StubbornAtom: $\lim_{x\to\infty} f'(x) = 0$; it says so right there in the question. – Javier Jun 11 at 15:36
In that case shouldn't it be 'the limit of the derivative that tends to zero'? I was referring to Michael Hardy's answer at the bottom. – StubbornAtom Jun 11 at 17:19

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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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! – Javier Jun 23 '12 at 19:39
Also, a nitpick: shouldn't it be $x > 0$ instead of $x \ge 0$? – Javier Jun 23 '12 at 19:39
@Javier Thanks. Including $x=0$ is not a problem but since this could distract the reader, I modified the answer. – Did Nov 16 '15 at 7:30

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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What is "nbhd"? – Hammerite Jun 23 '12 at 19:54
"Neighborhood". – Mat Jun 23 '12 at 19:56

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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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. – Henning Makholm Jun 23 '12 at 17:51

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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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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This is a good point, but should really be a comment, not an answer. – Noah Schweber Nov 16 '15 at 7:55
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. – Jae Young Lee Nov 17 '15 at 16:46