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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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    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.2018-01-29

6 Answers 6

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

$f'(c_x)=f(x+1)-f(x)$

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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    Do you really want an explanation? You have been warned not to post [duplicate answers](https://math.stackexchange.com/a/2626705/11619). The proper thing to do would have been to vote to close that as a dupe, and then possibly post this once.2018-01-30
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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.

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    @Chris See the answer linked in my first comment,2016-06-18
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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.

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    @Shai Covo I really like the idea of proving by contradiction. So if L>0, then there is some $N$ such that2015-01-23
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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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    @Kimo L'H can actually be applied for $\frac{anything}{\infty}$. Check the second case of this proof: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#General_proof2014-05-30
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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 $\lim_{n \to \infty}f^{'}(x)=0$.

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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). $

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    @Rhythmic Fistman: You may find the following very useful: http://58.20.53.14/ec/C404/Course/jiaoan/13-4.pdf2019-03-31