Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question

5 Answers 5

up vote 20 down vote accepted

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 $\limsup\limits_{x\to+\infty}\,f'(x)=+\infty$.

share|cite|improve this answer
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 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.

share|cite|improve this answer
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".

share|cite|improve this answer
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.

share|cite|improve this answer

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 for the statement of Barbalat's lemma and its applications in stability analysis).

share|cite|improve this answer
This is a good point, but should really be a comment, not an answer. – Noah Schweber Nov 16 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 at 16:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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