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

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    $\begingroup$ Hint: What is $\lim_{x \to \infty} \frac{f(x)}{x}$? $\endgroup$ – N. S. May 31 '11 at 5:14
  • $\begingroup$ @N.S. what is that ration give at the end? what is your point? $\endgroup$ – Guy Fsone Jan 29 '18 at 16:17
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    $\begingroup$ Surprised nobody has made this comment earlier - to see why you need to assume $\lim_{x\to \infty} f'(x)$ exists, consider the function $f(x) = \frac{\sin(x^2)}{x}$. Then $\lim_{x\to \infty} f(x) = 0$, yet $\limsup_{x\to \infty} f'(x) = 2$ and $\liminf_{x\to \infty} f'(x) = -2$, so $\lim_{x\to \infty} f'(x)$ does not exist. $\endgroup$ – Daniel Schepler Jan 29 '18 at 23:29

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.

  • $\begingroup$ 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 $\displaystyle\lim_{c \to \infty}f'(c) = 0$? $\endgroup$ – saurs May 31 '11 at 6:37
  • $\begingroup$ @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 $L<0$. $\endgroup$ – Shai Covo May 31 '11 at 6:52
  • $\begingroup$ OK, I've got it now. Thanks for the help. $\endgroup$ – saurs May 31 '11 at 7:47
  • $\begingroup$ What's wrong with swapping the order of the limits? $\endgroup$ – Rhythmic Fistman May 31 '11 at 18:30
  • $\begingroup$ @Rhythmic Fistman: Can you please be more specific? $\endgroup$ – Shai Covo May 31 '11 at 19:03

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.

  • $\begingroup$ You also need to assume that either $\lim \limits_\infty f$ or $\lim \limits_\infty f'$ exists. $\endgroup$ – Git Gud Apr 14 '14 at 20:35
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    $\begingroup$ @GitGud Not true, see the table in the article, or see the excerpt in this answer. $\endgroup$ – Bill Dubuque Apr 14 '14 at 21:00
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    $\begingroup$ Of course you're right, I had missed something. Thank you. $\endgroup$ – Git Gud Apr 14 '14 at 21:11
  • $\begingroup$ 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? $\endgroup$ – Ant Jun 30 '14 at 18:40
  • $\begingroup$ @Ant See the link in my prior comment. $\endgroup$ – Bill Dubuque Jun 30 '14 at 18:50

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 $\frac{xf(x)}{x^2}$ ;)

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    $\begingroup$ 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? $\endgroup$ – Pete L. Clark May 31 '11 at 15:17
  • $\begingroup$ :) we all miss simple things, eh? $\endgroup$ – N. S. May 31 '11 at 19:35
  • $\begingroup$ well, I can't speak for "all", but I know I do... $\endgroup$ – Pete L. Clark Jun 1 '11 at 16:54
  • $\begingroup$ @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? $\endgroup$ – otupygak May 30 '14 at 20:14
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    $\begingroup$ @Kimo L'H can actually be applied for $\frac{anything}{\infty}$. Check the second case of this proof: en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_proof $\endgroup$ – N. S. May 30 '14 at 20:25

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 $\lim_{n \to \infty}f^{'}(x)=0$.


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, $$ \mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} f(x,h) = \infty \neq 0 = \mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{x \to \infty } f(x,h). $$


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

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

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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jan 30 '18 at 13:00

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