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Not homework. I need this (or something similar) to solve 4. in this question.

Let $y:(a ,+\infty)\to \mathbb R$ be $C^1$.

Prove that $$\lim_{x\to +\infty}y(x)=\eta\text{ for some }\eta\in \mathbb R\implies\text{the following limit exists and } \lim_{x\to +\infty}y'(x)=0$$

Intuitively this is true because $\displaystyle \lim_{x\to +\infty}y(x)=\eta$ means that $y$ almost stops increasing or decreasing, so $\displaystyle \lim_{x\to +\infty}y'(x)=0$. But how to prove it?

I tried $$\lim_{x\to +\infty}y'(x)=0=\lim_{x\to +\infty} \lim_{h\to 0}\dfrac{y(x+h)-y(x)}h= \lim_{h\to 0}\lim_{x\to +\infty}\dfrac{y(x+h)-y(x)}h= \lim_{h\to 0}\dfrac{a-a}h=0,$$

but why I can change the order of the limits? If I can't even do that, how can I prove this?

After reading the threads Tyler provided, I now just need to prove that $\displaystyle \lim_{x\to +\infty}y'(x)$ exists? Please consider a suitable adaption to the linked question.

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antifb
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  • @Landscape I didn't think of that. Thank you for pointing it out. – antifb Oct 22 '13 at 18:37
  • @Landscape If it is false, what conditions do I need to add to make it true? Can you please make a suitable adaptation considering the linked question? – antifb Oct 22 '13 at 18:41
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    Here's a couple threads you might want to take a look at: Thread 1 and Thread 2, they answer your question. – Tyler Oct 22 '13 at 18:41
  • @Tyler Thank you, will do – antifb Oct 22 '13 at 18:42
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    For the question here, $$f(x) = \frac{\sin (x^4)}{1+x^2}$$ has $\lim\limits_{x\to \pm\infty} f(x) = 0$, is $C^\infty$, but $f'(x)$ is unbounded. – Daniel Fischer Oct 23 '13 at 11:12
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    For the linked question, you have that the derivative is a Lipschitz function of $y$, so if $\lim_{x\to\infty} y(x)$ exists and is in the domain of $f$, then $\lim_{x\to\infty} y'(x) = \lim_{x\to\infty} f(y(x)) = f(\lim_{x\to\infty}y(x))$ by the continuity of $f$ [you don't even need Lipschitz for that]. – Daniel Fischer Oct 23 '13 at 11:19
  • @DanielFischer Nice! (You should post the two comments as answers.) – Andrés E. Caicedo Oct 23 '13 at 14:45
  • @AndresCaicedo The first is just an explicit example of the phenomenon you have already described in your answer, feel invited to include it there, not worth an additional answer here. The second could indeed be made an answer to the other question. – Daniel Fischer Oct 23 '13 at 14:52

2 Answers2

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This is false without additional assumptions. If we suppose that $\lim_{x\to\infty}f'(x)$ exists, then an argument such as that of Elias shows that it must in fact be zero.

In the comments, some threads with different arguments proving the same are also suggested. Hmm... Elias deleted their answer, so let me include a nice approach (that is in one of the answers to the question this is a duplicate of): Suppose $\lim_{x\to\infty}f'(x)$ exists. We can then use L'Hôpital's rule (as formulated, for example, in Rudin's Principles of mathematical analysis) to conclude that $$ \lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{e^x f(x)}{e^x}=\lim_{x\to\infty}\frac{e^x(f(x)+f'(x))}{e^x}=\lim_{x\to\infty}{f(x)+f'(x)}, $$ where the third equality is by L'Hôpital's rule, and its application is justified since the last limit exist, by assumption.

However, if the assumption that $\lim_{x\to\infty}f'(x)$ exists is not given, we can produce counterexamples.

To see this, imagine $f$ is a piecewise linear function that is zero except in tiny neighborhoods of $x=n$ for all $n$. For each $n$, $f(x)\ne0$ for $x\in (n-(1/n^2), n+(1/n^2))$, where it is a straight line from $(n-1/n^2,0)$ to $(n,1/n)$, and then a straight line from $(n,1/n)$ to $(n+1/n^2,0)$. Note that $f(x)\to0$ as $x\to\infty$, since the "peaks" have heights decreasing to $0$. However, the lines around each integer are getting steeper.

This $f$ is not differentiable everywhere, but there are standard methods that allow us to redefine $f$ near the points where $f'$ is not defined, so that the redefined function is smoother and $f'$ exists everywhere, see for example this question (essentially, we "round" the corners). We still have $f\to0$ but now $f'$ exists everywhere, and $\lim_{x\to\infty}f'(x)$ does not exist, because for each integer $n>0$, near $x=n$, $f'$ takes values $n$ and $-n$.

To increase their visibility, let me add two comments by Daniel Fisher:

An explicit example of a $C^\infty$ function $f$ such that $\lim_{x\to\infty} f(x)$ exists but $\lim_{x\to\infty}f'(x)$ does not is $$ f(x)=\frac{\sin(x^4)}{1+x^2}. $$ One can arrive at this example by starting with something like $g(t)=t^2\sin(1/t)$ which is differentiable at $0$ but has discontinuous derivative at $0$, and modifying it so the behavior occurs at $+\infty$ rather than $0$ (a small further change is added to ensure the function is defined everywhere, even at $0$).

Fisher further adds: For the linked question, you have that the derivative is a Lipschitz function of $y$, so if $\lim_{x\to\infty}y(x)$ exists and is in the domain of $f$, then $$\lim_{x\to\infty}y'(x)=\lim_{x\to\infty}f(y(x))=f(\lim_{x\to\infty}y(x))$$ by the continuity of $f$, and you don't even need Lipschitz for that.

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Andrés E. Caicedo
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Here is another easy way to prove that $\lim f'(y)=0$ if this limit exists.

Suppose : $\displaystyle\lim_{y\to +\infty} f'(y)=\lambda\neq 0$.

Without loss of generality, we can assume $\lambda>0$ (you can replace $f$ with $-f$ in the other case).

Then there is a $b>a$ such that $f'(x)>\dfrac\lambda 2$ for all $x\geqslant b$.

This leads you to :

$$f(x)=f(0)+\int_0^x f' \geqslant f(0) + \frac\lambda 2 x,\ (\text{if}\ x\geqslant b)$$

But $f(0)+\dfrac\lambda 2 x$ tends to $+\infty$ as $x$ tends to $+\infty$ .

This is contradictory, so $\displaystyle\lim_{y\to +\infty} f'(y)=0$

Philippe Malot
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