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Suppose that $\lim_{x\to \infty} f'(x) = a$. Is it true that $\lim_{x\to \infty} {f(x)\over x} = a$

If so, can you prove it? Thanks!

pbk83
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3 Answers3

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In case L'Hopital's Rule is not allowed :

Let $\epsilon > 0$, Then there exist larg enough $M >0 $ such that $ a-\epsilon \leq f'(x) \leq a+ \epsilon $, for all $ M\leq x $. Now Apply mean value theorem for $f$ on $[M , x ]$ then there exist $y_x \in [M, x]$ such that $$ a- \epsilon \leq f'(y_x) = \frac{f(x) - f(M)}{ x - M} \leq a+ \epsilon $$

Now letting $x \to + \infty$ we get $$ a - \epsilon \leq \limsup_{x \to + \infty} \frac{f(x)}{ x} \leq a+ \epsilon $$ for all $\epsilon > 0$ which implies $ \limsup_{x \to + \infty} \frac{f(x)}{ x} =a$, similarly $ \liminf_{x \to + \infty} \frac{f(x)}{ x} = a$

Red shoes
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  • When you take limits as $x\to\infty$ what happened to $M$ and $f(M) $? Also you should get your last equation in terms of limsup and liminf and not in form of limit as you don't know beforehand that it exists. These objections are not too difficult to handle but it is better to address them and fix your answer accordingly. – Paramanand Singh Jun 17 '17 at 12:09
  • The behavior of $\frac{f(x) - f(M)}{ x - M}$ and $ \frac{f(x)}{x} $ are same, as $x \to \infty$ and this clear since $M$ is fixed ! $$\frac{f(x)}{x} = \frac{f(x) - f(M)}{ x - M} \times \frac{x-M}{x} + \frac{M}{x}$$

    For limsup you are right! it is better to be pointed out !

     – Red shoes Jun 17 '17 at 12:29
  • Exactly! It's great that you understood my point. Please update your answer accordingly. – Paramanand Singh Jun 17 '17 at 12:32
  • Pending +1 delivered. – Paramanand Singh Jun 17 '17 at 12:37
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Let's assume $f'(x)\rightarrow0$ as $x\rightarrow\infty$. Then, there's a $x_\epsilon$ so that $|f'(x)|<\epsilon$ for $x>x_\epsilon.$ We can estimate $$|f(x)|\le|f(x_{\epsilon/2})|+|f(x)-f(x_{\epsilon/2})|=|f(x_{\epsilon/2})|+|(x-x_{\epsilon/2})f'(\xi)|$$ with some $\xi\in[x_{\epsilon/2},x]$ (mean value theorem), so $|f'(\xi)|<\epsilon/2.$ The last summand is $<x\,\epsilon/2$, so $$|f(x)/x|<\frac{|f(x_{\epsilon/2})|}{x}+\epsilon/2<\epsilon$$ for $x>2\,\frac{|f(x_{\epsilon/2})|}{\epsilon}.$
For the general case, consider $f(x)-ax$ instead of $f(x).$
To explain the downvotes: the original answer included unnecessarily heavy machinery, the reasoning was only justified with Lebesgue integrals and absolute continuity, being overkill for such a relatively simple statement.

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    How did you get that estimation? –  Jun 17 '17 at 11:21
  • This is a comment from @csstudent:

    $f$ might have a derivative $f'$ but $f'$ might not be integrable.

    For example this one.

     – Red shoes Jun 17 '17 at 11:28
  • The problem is not about not liking Lebesgue integrals. Even if you use Henstock integrals the first proof is wrong, because it is assuming that $|f'|$ is integrable, which is not always the case. – OR. Jun 17 '17 at 11:44
  • I think this proof (first proof ) is totally fine if we interpreted those integrals as Upper Remman integral, Note that $f'$ is eventually bounded @MlazhinkaShungGronzalezLeWy – Red shoes Jun 17 '17 at 11:52
  • @Ashkan Boundedness has nothing to do. You can have f' bounded and still not integrable. Those upper Riemann integrals would be infinite and the last inequality would be $x>\infty$ – OR. Jun 17 '17 at 11:53
  • @Mlazhinka Shung Gronzalez LeWy Under our assumptions, $f'$ is bounded for large enough arguments. Show me a bounded measurable function that is not integrable, please. If $f'$ is bounded, $f$ is absolutely continuous, and $f(x)-f(x_0)=\int^x_{x_0}f'(t),dt,$ at least in the sense of a Lebesgue integral. Of course, that's rather heavy machinery for such an elementary problem. –  Jun 17 '17 at 11:54
  • You always have Upper integral @MlazhinkaShungGronzalezLeWy – Red shoes Jun 17 '17 at 11:54
  • https://en.wikipedia.org/wiki/Volterra's_function – OR. Jun 17 '17 at 11:55
  • @Mlazhinka Shung Gronzalez LeWy That derivative is not Riemann-integrable. –  Jun 17 '17 at 11:57
  • Exactly the point – OR. Jun 17 '17 at 11:58
  • Using integrals is unnecessary when the problem is easily solved by mean value theorem. Especially the use of Lebesgue theory of integration and things like absolute continuity are an overkill. One should strive for simplicity. – Paramanand Singh Jun 17 '17 at 12:30
  • @Paramanand Singh You are absolutely correct, I'll edit my solution to put the more useful answer first. –  Jun 17 '17 at 14:55
  • BTW, just curious are you a professor? – Paramanand Singh Jun 17 '17 at 18:55
  • @Paramanand Singh In the sense of most countries, I was a professor, i.e. I taught mathematics at a university. But that was long ago, last millenium. For the last seventeen years or so, I'm a software engineer, and mathematics is only a pastime. –  Jun 17 '17 at 19:03
  • After your edit I gave my pending +1. – Paramanand Singh Jun 17 '17 at 19:08
  • @Paramanand Singh Ah, that were you, thanks! I'm afraid the downvotes were correct, though. This here is not only about being right, the answer should be useful for the OP, too. –  Jun 17 '17 at 19:43
  • I did not downvote. I can see you have two downvotes and 3 upvote and last upvote is mine. Normally I downvote answers which are incorrect (perhaps with big mistakes). – Paramanand Singh Jun 18 '17 at 01:02
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HINT

Let assume $a \gt 0$. Then, because $\lim_{x\to \infty} f'(x) = a \gt 0$ there is $M \gt 0$ such that $f'(x) \gt 0$ for all $x \ge M$. Therefore $f$ is strictly increasing on $[M, +\infty)$ and unbounded (otherwise $\lim_{x\to \infty} f'(x) =0$). It follows that $\lim_{x\to \infty} f(x) = +\infty$ and we can apply L'Hospital to $\lim_{x\to \infty} {f(x)\over x}$

Now, let's assume $a=0$. Let $g(x) = f(x) + x$. Then $\lim_{x\to \infty} g'(x)=1$ therefore $\lim_{x\to \infty} {g(x)\over x}=1$ and from here $\lim_{x\to \infty} {f(x)\over x}=0$

  • There is no need to check the unboundedness of $f$. For L'Hospital's Rule to be applicable it is sufficient that the denominator tends to $\pm\infty$ and one does not need to check the numerator. – Paramanand Singh Jun 17 '17 at 12:28