By saurs

2011-05-31 05:09:40 8 Comments

I've been trying to solve the following problem:

Suppose that $f$ and $f'$ are continuous functions on $\mathbb{R}$, and that $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to\infty}f'(x)$ exist. Show that $\displaystyle\lim_{x\to\infty}f'(x) = 0$.

I'm not entirely sure what to do. Since there's not a lot of information given, I guess there isn't very much one can do. I tried using the definition of the derivative and showing that it went to $0$ as $x$ went to $\infty$ but that didn't really work out. Now I'm thinking I should assume $\displaystyle\lim_{x\to\infty}f'(x) = L \neq 0$ and try to get a contradiction, but I'm not sure where the contradiction would come from.

Could somebody point me in the right direction (e.g. a certain theorem or property I have to use?) Thanks


@Guy Fsone 2018-01-29 16:14:21

The Mean value theorem,reveals that, for each $x$ there exists $c_x\in (x,x+1)$ such that


Hence since, $c_x\to\infty$ as $x\to\infty$ we have

$$\lim_{x\to \infty}f'(x)=\lim_{x\to \infty}f'(c_x) =\lim_{x\to \infty}[f(x+1)-f(x)] =0 $$

@Guy Fsone 2018-01-30 11:42:10

Who ever down voted this must explain. It is helpful to understand what went wrong

@Jyrki Lahtonen 2018-01-30 13:00:42

Do you really want an explanation? You have been warned not to post duplicate answers. The proper thing to do would have been to vote to close that as a dupe, and then possibly post this once.

@Bill Dubuque 2011-06-02 19:35:33

Apply a L'Hospital slick trick: $\, $ if $\rm\ f + f\,'\!\to L\ $ as $\rm\ x\to\infty\ $ then $\rm\ f\to L,\ f\,'\!\to 0,\, $ since

$$\rm \lim_{x\to\infty}\ f(x)\ =\ \lim_{x\to\infty}\frac{e^x\ f(x)}{e^x}\ =\ \lim_{x\to\infty}\frac{e^x\ (f(x)+f\,'(x))}{e^x}\ =\ \lim_{x\to\infty}\, (f(x)+f'(x))\qquad $$

This application of L'Hôpital's rule achieved some notoriety because the problem appeared in Hardy's classic calculus texbook A Course of Pure Mathematics, but with a less elegant solution. For example, see Landau; Jones: A Hardy Old Problem, Math. Magazine 56 (1983) 230-232.

@Git Gud 2014-04-14 20:35:47

You also need to assume that either $\lim \limits_\infty f$ or $\lim \limits_\infty f'$ exists.

@Bill Dubuque 2014-04-14 21:00:07

@GitGud Not true, see the table in the article, or see the excerpt in this answer.

@Git Gud 2014-04-14 21:11:22

Of course you're right, I had missed something. Thank you.

@Ant 2014-06-30 18:40:43

Wouldn't you need to assume that $\lim e^x f(x) = \infty$? Because otherwise you can't apply L'Hopital rule.. Or am I missing something?

@Bill Dubuque 2014-06-30 18:50:21

@Ant See the link in my prior comment.

@Rusty Nail 2016-02-22 23:17:44

@Bill Dubuque - so as you approach Infinity, the Derivative should return Zero? EG: delta *= Derivative | where Derivative is Zero before it is Infinity? Or should this be One instead of Zero?

@Bill Dubuque 2016-06-18 23:35:18

@Chris See the answer linked in my first comment,

@Shai Covo 2011-05-31 05:27:27

Hint: If you assume $\lim _{x \to \infty } f'(x) = L \ne 0$, the contradiction would come from the mean value theorem (consider $f(x)-f(M)$ for a fixed but arbitrary large $M$, and let $x \to \infty$).

Explained: If the limit of $f(x)$ exist, there is a horizontal asymptote. Therefore as the function approaches infinity it becomes more linear and thus the derivative approaches zero.

@saurs 2011-05-31 06:37:24

Ah ok, so $\displaystyle\lim_{x\to\infty}\frac{f(x) - f(M)}{x - M} = 0$, right? (Which I suppose is what user9176 was implying.) So just to make sure I'm clear on this, if we take $\frac{f(x) - f(M)}{x - M} = f'(c)$ for some $c \in (M, x)$ then as $x \to \infty$ the left-hand side goes to $0$. And since we take $M$ arbitrarily large does it follow that $c \to \infty$, and hence $\displaystyle\lim_{c \to \infty}f'(c) = 0$?

@Shai Covo 2011-05-31 06:52:08

@sarus: I suggest proving the result by contradiction, that is by assuming $\lim _{x \to \infty } f'(x) = L \ne 0$ (as you originally tried). It may be comfortable for you to split into the cases $L>0$ and $L<0$.

@saurs 2011-05-31 07:47:01

OK, I've got it now. Thanks for the help.

@Rhythmic Fistman 2011-05-31 18:30:38

What's wrong with swapping the order of the limits?

@Shai Covo 2011-05-31 19:03:53

@Rhythmic Fistman: Can you please be more specific?

@Rhythmic Fistman 2011-05-31 21:21:19

@Shai Covo: I mean naively changing the order of the x limit and the limit implicit* in the derivative. The MVT seems a little heavy handed for this result (no offence), I was hoping for a more minimal proof. * I'm also hoping to be reminded why you can't do this in general.

@Shai Covo 2011-05-31 21:53:11

@Rhythmic Fistman: Your intention is still unclear. You may wish to elaborate on your idea in a new answer, starting with "this is too long to be a comment" etc. (or, of course, elaborate in a new comment below). Anyway, take into account that MVT is a very elementary tool, as far as rigorous proofs are concerned.

@Rhythmic Fistman 2011-06-02 09:15:19

This is question about the limit of a derivate, but the derivate is itself a limit - so you have two limits. If you exchange their order, the result is obviously zero. Is this a simpler proof or is the limit exchange step wrong or difficult to justify?

@Shai Covo 2011-06-02 10:12:02

@Rhythmic Fistman: I finally understand your question. In general, such steps are either wrong or difficult to justify. I'll consider this specific case (but not soon) and let you know if I find something useful to say.

@Shai Covo 2011-06-02 15:03:46

@Rhythmic Fistman: See my new answer...

@math101 2015-01-23 07:07:34

@Shai Covo I really like the idea of proving by contradiction. So if $L>0$, then there is some $N$ such that

@Beni Bogosel 2011-05-31 09:01:28

You know that $\lim_{x \to \infty}f(x)$ and $\lim_{x \to \infty}f^{'}(x)$ exists. Then by Lagrange's theorem there exists $c_n \in (n,n+1)$ such that $f(n+1)-f(n)=f^{'}(c_n)$ Taking the limit as $n \to \infty$ you get that $\lim_{n \to \infty}f'(c_n)=0$. Since the limit exists, and there exists a sequence for which the limit of the function is $0$ it follows that $\lim_{n \to \infty}f^{'}(x)=0$.

@Shai Covo 2011-06-02 15:04:35

This is in response to an interesting observation made by Rhythmic Fistman in a comment below my (first) answer. We suppose that $\lim _{x \to \infty } f'(x) = L$ for some $L \in \mathbb{R}$. Then, from the definition of the derivative, $$ L = \mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}. $$ As Rhythmic Fistman observed, naively changing the order of the limits gives rise to the equality $$ L = \mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{x \to \infty } \frac{{f(x + h) - f(x)}}{h} = 0, $$ where the last equality follows from the assumption that $\lim _{x \to \infty } f(x)$ exists (finite). ``Hence'' the desired result $\lim _{x \to \infty } f'(x) = 0$. However, as the following counterexample shows, this procedure is not allowed in principle. Define a two-variable function $f$ by $f(x,h)=xe^{-|h|x}$. Analogously to the case in the original question (where the role of $f(x,h)$ is played by $\frac{{f(x + h) - f(x)}}{h}$), $$ \mathop {\lim }\limits_{x \to \infty } f(x,h) = \mathop {\lim }\limits_{x \to \infty } xe^{ - |h|x} = 0, $$ for any $h \neq 0$. Hence, $$ \mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{x \to \infty } f(x,h) = 0. $$ Also, for any $x \in \mathbb{R}$, $$ \mathop {\lim }\limits_{h \to 0} f(x,h) = \mathop {\lim }\limits_{h \to 0} xe^{ - |h|x} = x $$ (this is analogous to the case in the original question, where $f'$ is assumed continuous on $\mathbb{R}$). Hence, $$ \mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} f(x,h) = \infty \neq 0 = \mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{x \to \infty } f(x,h). $$

@Rhythmic Fistman 2011-06-08 21:15:42

Thanks, good counter example, I think I get it. So in general limits don't commute - are there any conditions under which they do?

@Shai Covo 2011-06-09 14:34:16

@Rhythmic Fistman: For nontrivial conditions, this paper may be relevant:

@Shai Covo 2011-06-09 14:41:37

@Rhythmic Fistman: You may find the following very useful:

@N. S. 2011-05-31 15:10:37

To expand a little on my comment, since $\lim_{x \to \infty} f(x) = L$, we get

$$\lim_{x \to \infty} \frac{f(x)}{x} =0 \,.$$

But also, since $\lim_{x \to \infty} f'(x)$ exists, by L'Hospital we have

$$\lim_{x \to \infty} \frac{f(x)}{x}= \lim_{x \to \infty} f'(x) \,.$$

Note that using the MTV is basically the same proof, since that's how one proves the L'H in this case....

P.S. I know that if $L \neq 0$ one cannot apply L'H to $\frac{f(x)}{x}$, but one can cheat in this case: apply L'H to $\frac{xf(x)}{x^2}$ ;)

@Pete L. Clark 2011-05-31 15:17:11

This a nice answer. As for the P.S., rather than your "cheat", it seems more natural to just replace $f(x)$ with $f(x) - L$, doesn't it?

@N. S. 2011-05-31 19:35:13

:) we all miss simple things, eh?

@Pete L. Clark 2011-06-01 16:54:36

well, I can't speak for "all", but I know I do...

@otupygak 2014-05-30 20:14:04

@N.S.Sorry if this seems like a silly question, but to use L'Hopital doesn't the fraction $f(x)/g(x)$ have to be either of $0/0 \ , \ \infty / \infty \ or \ -\infty / -\infty$ in this case as $x$ goes to infinity $f(x)$ goes to 0 while $x$ goes to $\infty$ so you have a fraction of the form $0 / \infty$?? can you still use l'hopital for this?

@N. S. 2014-05-30 20:25:13

@Kimo L'H can actually be applied for $\frac{anything}{\infty}$. Check the second case of this proof:

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