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I have this problem:

Let $c\in \mathbb{R}$. If $\int_c^\infty f(x)dx$ converges, then $$\lim_{x\to \infty} f(x)$$ exist and is $0$. Moreover, if $f$ is monotonic, $\lim_{x\to \infty} xf(x)$ exist and is equal to $0$.

Although I can't assume that $f$ is continuous, I have tihis question:

If $F$ is a function with continuous first derivative, such that $$\lim_{x\to \infty}F(x)=L,$$ for some $L\in\mathbb{R}$. Is it true that $$\lim_{x\to \infty} F'(x)=0\;?$$

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  • $\begingroup$ @leo: first work out it for the case when $c=0$. It is simpler. $\endgroup$ – user9413 May 31 '11 at 5:14
  • $\begingroup$ @Please see problem 1.5.15 in Kaczor's Problems in Mathematical Analysis, Vol III, Integration. $\endgroup$ – user9413 May 31 '11 at 5:34
  • $\begingroup$ Then is necessary the uniform continuity of $f$. Thanks @Chandru. $\endgroup$ – leo May 31 '11 at 6:30
  • $\begingroup$ In the course notes, the problem is as I write here. $\endgroup$ – leo May 31 '11 at 6:33
  • $\begingroup$ @leo: No i just gave you a similar problem. $\endgroup$ – user9413 May 31 '11 at 6:38
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André already gave you a very simple example, where $\lim\limits_{x \to \infty } F(x) = 0$ but $\underset{x \to \infty }{\lim \sup}\; F'(x) = 2$ and $\underset{x \to \infty }{\lim \inf}\; F'(x) = -2$ (since $F'(x) = 2\cos (x^2 ) - \frac{{\sin (x^2 )}}{{x^2 }}$). Modifying $F$ to $F(x) = \frac{{\sin (x^3 )}}{x}$, we moreover get an example with $\underset{x \to \infty }{\lim \sup}\; F'(x) = \infty$ and $\underset{x \to \infty }{\lim \inf}\; F'(x) = -\infty$ (since $F'(x)=3x\cos (x^3 ) - \frac{{\sin (x^3 )}}{{x^2 }}$). Obviously, in both examples, $F$ is not a monotone function.

Let's now give an example of a (continuously differentiable) monotone increasing function $F$ with $\lim\limits_{x \to \infty } F(x) = 1$, for which, nevertheless, $\underset{x \to \infty }{\lim \sup}\; F'(x) = \infty$ (so, in particular, $F'(x)$ does not tend to $0$ as $x \to \infty$). We first define a continuous function $f:[0,\infty) \to [0,\infty)$ as follows: for each positive integer $n$, $f(n)=0$, $f$ is linearly increasing on the interval $\left[n,n+\frac1{2^{n+1}}\right]$, $f\left(n+\frac1{2^{n+1}}\right) = n$, $f$ is linearly decreasing on the interval $\left[n+\frac1{2^{n+1}},n+\frac1{2^n}\right]$, and $f\left(n+\frac1{2^n}\right)=0$; for any other $x$, we define $f(x)=0$. Then we have $$ \int_0^\infty {f(x)\,\mathrm dx} = \sum_{n = 1}^\infty {\int_n^{n + 2^{ - n} } {f(x)\,\mathrm dx} } = \sum_{n = 1}^\infty {\frac{n2^{-n}}{2}} = \frac12\sum_{n = 1}^\infty \frac{n}{2^n} = 1. $$ Now we define the function $F$ by $F(x) = \int_0^x {f(t)\,\mathrm dt}$. Then, $$ \lim _{x \to \infty } F(x) = \lim _{x \to \infty } \int_0^x {f(t)\,\mathrm dt} = \int_0^\infty {f(t)\,\mathrm dt} = 1. $$ On the other hand, by the Fundamental Theorem of Calculus, $F'(x) = f(x)$. Since, for any positive integer $n$, $f\left(n+\frac1{2^{n+1}}\right) = n$, we have $$ \underset{x \to \infty }{\lim \sup}\; F'(x) = \underset{x \to \infty }{\lim \sup}\; f(x) = \infty . $$

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It's true iff $\rm\ lim\ F\,'$ exists. Andre gave a counterexample if $\rm\ lim\ F\,'$ does not exist. Conversely:

Theorem $\ $ If $\rm\,\ F + F\,'\!\to L\ $ as $\rm\ x\to\infty\ $ then $\rm\ F\to L,\ F\,'\!\to 0,\,\ $ by this L'Hôpital slick trick:

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

The above employs a slightly generalized form of L'Hospital's rule mentioned here.

This folklore L'Hospital trick is somewhat notorious 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. Below is a table of the various possibilities, with examples, where FTE = Fails To Exist. L'Hospitable table L'Hospitable table explanation

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Let $$F(x)=\frac{\sin(x^2)}{x}$$ Then $\lim_{x\to\infty} F'(x)$ does not exist.

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  • $\begingroup$ is this function continuous at $0$. $\endgroup$ – user9413 May 31 '11 at 8:04
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    $\begingroup$ @Chandru: Continuity at $0$ isn't relevant to the question, but if you set $F(0)=0$, the function is continuously differentiable on $\mathbb{R}$. $\endgroup$ – Jonas Meyer May 31 '11 at 8:22
  • $\begingroup$ @Jonas Meyer: Thanks for responding to the question. When first writing the answer, I had $\sqrt{x^2+1}$ at the bottom, then decided to replace with $x$, and not to bother with specifying that $F(0)=0$. Wouldn't matter anyway, even if there were a natural non-removable discontinuity at $0$, one could splice over it in an infinitely differentiable way. $\endgroup$ – André Nicolas May 31 '11 at 9:29

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