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Suppose $f:\mathbb{R}\to\mathbb{R}$ is everywhere differentiable, and suppose that $\lim_{x\to\infty}f(x)$ and $\lim_{x\to\infty}f^{\prime}(x)$ both exist. I am trying to prove that the latter limit is necessarily $0$. I have the following argument, but I'm not sure if it's completely sound.

Since $f$ is differentiable everywhere, we can apply the Mean Value Theorem to $f$ on $[x,x+1]$ for all relevant $x$. This guarantees an $\alpha_{x}\in(x,x+1)$ such that $$f^{\prime}(\alpha_{x}) = \frac{f(x+1)-f(x)}{x+1-x} = f(x+1)-f(x).$$ Now, the limit as $x\to\infty$ of the right-hand side of this expression must be $0$, since $\lim_{x\to\infty}f(x)$ exists by assumption (and must equal $\lim_{x\to\infty}f(x+1)$). On the left hand side, we notice that $\alpha_{x}\to\infty$ as $x\to\infty$, since $\alpha_{x}>x$ always, so that: \begin{eqnarray*} 0 & = & \lim_{x\to\infty}[f(x+1)-f(x)]\\ & = & \lim_{x\to\infty}f^{\prime}(\alpha_{x})\\ & = & \lim_{y\to\infty}f^{\prime}(y), \end{eqnarray*} proving the result.

I took inspiration for this argument from other sources which use the same trick of "use the Mean Value Theorem to introduce a quantity $\alpha_{x}$ which we have some bounds on, then take limits". However, this style of argument seems dodgy to me: we haven't actually defined a function $\alpha$ to take the limit of as $x\to\infty$, and it's not clear to me that defining such a function is always possible. For example, we can't just say "take the least such value and call it $\alpha_{x}$", because we haven't shown that there will always be a least such value.

Here are my questions:

  • In the above, where have we used the fact that $\lim_{x\to\infty}f^{\prime}(x)$ exists? This is an important assumption: consider for example the function $x\mapsto\sin{(x^{2})}/x$. My guess is that it's used in the last line, where we must assume this fact to use the chain rule, but I'd like confirmation of this.

  • Does the "$\alpha_{x}$ trick" require something like the Axiom of Choice in general? In particular, the thing which makes me slightly anxious about just saying "choose an $\alpha_{x}$ for every $x$" is that we have to make (uncountably) infinitely many "choices", and we have no prescribed method of doing this. EDIT: It turns out this has been answered in other questions on this site, see link in the comments below.

EDIT: Note that the first question is different to others on related topics because here I am asking very specifically about this argument and why it works.

Will R
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  • You have to suppose that $\lim_{x\to \infty} f'(x)$ exist for the following reason: You want to calculate $\lim_{x\to \infty} f'(\alpha_x)$ and to conclude that $\lim_{x\to \infty} f'(x)=0$. Note that the for the calculation of $\lim_{x\to \infty} f'(\alpha_x) =0$ you don't need that $\lim_{x\to \infty} f'(x)$ exist. You need this assumption to conclude that $\lim_{x\to \infty} f'(x)=0$ because of uniqueness of limit. Your argument with $\alpha_x$ is totally rigorous. You are choosing an $a_x$ for each $x\in \mathbb R$ and define the map $x\mapsto a_x$. – Hugo Mar 20 '16 at 14:51
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    concerning choice in MVT: see the question mean-value-theorem-and-the-axiom-of-choice – user251257 Mar 20 '16 at 14:57
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    Old topics http://math.stackexchange.com/questions/44009/does-finiteness-of-lim-x-to-inftyfx-and-lim-x-to-inftyfx-imply/ and http://math.stackexchange.com/questions/42277/limit-of-the-derivative-of-a-function-as-x-goes-to-infinity/? – rtybase Mar 20 '16 at 15:39
  • @user251257: Thanks, I'll check that out. – Will R Mar 20 '16 at 15:46

3 Answers3

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As for your first question: the mean value theorem only guarantees the existence of some $\alpha_x$ with the given property, but you have no control where $\alpha_x$ is exactly found in the interval. The fact that $f^\prime$ converges allows you to to conclude that this does not matter, any sequence $x_n$ converging to $\infty$ will have the property $f^\prime(x_n)\rightarrow \lim_{x\rightarrow\infty} f^\prime(x)$, in particular the one you get using the MVT.

I don't get what you want to know with your second question. The proof you found is rigorous.

Thomas
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  • I have amended my second question. As for your answer to my first question, I still don't quite understand: where exactly does the argument break down for the function $f:x\mapsto\sin{(x^{2})}/x$? I seem to be getting $0 = \lim_{x\to\infty}f^{\prime}(\alpha_{x})$. What stops us from concluding that $\lim_{y\to\infty}f^{\prime}(y)=\lim_{x\to\infty}f^{\prime}(\alpha_{x})$ in this case? – Will R Mar 20 '16 at 14:58
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    @WillR: you have $f'(x) = -\sin(x^2)/x^2 + 2\cos(x^2)$, the limit for $x\to\infty$ does not exists. – user251257 Mar 20 '16 at 15:01
  • I know the limit does not exist. My point is that the argument seems to show that the limit not only does exist, but also that its value is $0$. – Will R Mar 20 '16 at 15:47
  • @WillR: The argument assumes that the limit exists, which is not satisfied. – user251257 Mar 20 '16 at 15:52
  • Where exactly does the argument assume this? The argument seems to say "$f^{\prime}(\alpha_{x})=f(x+1)-f(x)$ and the right-hand side tends to $0$ as $x\to\infty$". Why is this not a valid proof that $f^{\prime}(\alpha_{x})\to0$ as $x\to\infty$? Is the counterargument something like: for this $f$, we can find $\beta_{x}\in(x,x+1)$ such that $f^{\prime}(\beta_{x})\not\to0$ as $x\to\infty$? – Will R Mar 20 '16 at 16:03
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    @WillR if you say $\lim{x\rightarrow b} g(x) = a$ then this means two things: (i) $g(x_n)$ converges as $x_n$ tends to b for every sequence approaching $b$, and (ii) the limit is $a$. If you alread know it converges the limit is uniquely determined and you only have to consider one such sequence $x_n\rightarrow b$ to figure out what $a$ is. If you do not know it converges you have to show it tends to $a$ for every possible sequence. In case of your example it would break down because you can find other sequences which tend to some other (or no) value. – Thomas Mar 20 '16 at 16:27
  • Okay, I think we're on the same track now. The only slight discrepancy is that $\alpha_{x}$ isn't a sequence, it's a function, but I think this can be cleared up by rephrasing the whole argument in terms of an arbitrary sequence $(x_{n})$ that blows up as $x\to\infty$. – Will R Mar 20 '16 at 16:39
  • @WillR the mean value allows you to choose such an $\alpha_x$. You have, however no control where to find if you don't know the function, nor is it uniquely determined. There can be many (in fact, infinitely many) possible choices, or you may have only one such choice. So it is rather difficult to think of it as a function. You can, however, choose at least one such value in, say, $[n,n+1]$ and call it $\alpha_n$. That will do for your argument (because then $\alpha_n$ will be larger than $n$, hence tend to $\infty$, and cause you know the limit you can use that sequence to calculate it) – Thomas Mar 20 '16 at 17:04
  • What makes choosing a value for $\alpha_{n}$ any different from choosing a value for $\alpha_{x}$? – Will R Mar 20 '16 at 17:52
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For every $n$, by the mean value theorem, you have $f(n+m)-f(n)=mf'(x_{n,m})$. Since $lim_{x\rightarrow +\infty}f(x)=l$ exists, there exists $N_n$ such, $m>N_n$ implies $\mid f(n+m)-l\mid <1$, so $\mid f(n+m)-f(n)\mid =m\mid f'(x_{n,m})\mid< \mid l|+1+\mid f(n)\mid$ we deduce that

$\mid f'(x_{n,m})\mid \leq {{1+\mid l\mid +\mid f(n)\mid}\over m}$. This implies that for every $n>0,$ for every $c>0$ there exists $x_{n,m_0}=u_n>n$ such that $\mid f'(u_n)\mid <{1\over c}$. Remark that the limit of $u_n$ is $+\infty$ and the limit of $f'(u_n)$ is zero. Since $lim_{x\rightarrow+\infty}f'(x)$ exists, we deduce that this limit is zero.

Tsemo Aristide
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The only property of $\alpha_{x}$ which matters here is that $x<\alpha_{x}<x+1$ and we don't need to know details of how $ \alpha_{x}$ depends on $x$. We know that $f(x+1)-f(x)$ tends to $0$ and hence given any $\epsilon>0$ there exists an $M >0$ such that $$|f(x+1)-f(x)|<\epsilon\tag{1}$$ whenever $x>M$. Futhere we know that $f'(x)$ tends to $L$ as $x\to \infty$ so that there is a number $N>0$ such that $$L - \epsilon <f'(x)<L +\epsilon\tag{2}$$ whenever $x>N$. It is now obvious by mean value theorem that there is an $\alpha_{x}\in(x,x+1)$ such that $$-\epsilon<f'(\alpha_{x})<\epsilon\tag{3}$$ whenever $x>M$. If $L\neq 0$ then we can't have both inequalities $(2)$ and $(3)$ simultaneously for $\epsilon = |L|/2$ and $x > \max(M,N)$. Hence $L=0$.

Paramanand Singh
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