# Limit of the derivative of a function as x goes to infinity

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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2
Hint: What is $\lim_{x \to \infty} \frac{f(x)}{x}$? – N. S. May 31 '11 at 5:14

## 5 Answers

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 $x \to \infty$).

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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 $\displaystyle\lim_{c \to \infty}f'(c) = 0$? – saurs May 31 '11 at 6:37
@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 $L<0$. – Shai Covo May 31 '11 at 6:52
OK, I've got it now. Thanks for the help. – saurs May 31 '11 at 7:47
What's wrong with swapping the order of the limits? – Rhythmic Fistman May 31 '11 at 18:30
@Rhythmic Fistman: Can you please be more specific? – Shai Covo May 31 '11 at 19:03
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It follows by a L'Hôpital slick trick: $\$ if $\rm\ f + f\:\:'\!\to L\$ as $\rm\ x\to\infty\$ then $\rm\ f\to L,\ f\:\:'\!\to 0\:,\$ since

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

This L'Hospital rule trick achieved some noteriety due to the fact that 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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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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Thanks, good counter example, I think I get it. So in general limits don't commute - are there any conditions under which they do? – Rhythmic Fistman Jun 8 '11 at 21:15
– Shai Covo Jun 9 '11 at 14:34
– Shai Covo Jun 9 '11 at 14:41

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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1
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? – Pete L. Clark May 31 '11 at 15:17
:) we all miss simple things, eh? – N. S. May 31 '11 at 19:35
well, I can't speak for "all", but I know I do... – Pete L. Clark Jun 1 '11 at 16:54
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