A function $ f:\mathbb{R}_{\ge 0}\longrightarrow\mathbb{R} $ has the property that $ \lim_{x\to\infty } \frac{1}{x^2}\int_0^x f(t)dt=1. $ a) Give an example of what $ f $ could be if it's continuous and $ f/\text{id.} $ doesn't have a limit at $ \infty . $ b) Prove that if $ f $ is nondecreasing then $ f/\text{id.} $ has a limit at $ \infty , $ and determine it.
Problem
Source: Romanian National Olympiad 2017, grade xii, p.4
Tags: function, Asymptote, real analysis
26.08.2019 13:11
i dont get this qn because $ 1= \lim_{x\to\infty } \frac{1}{x^2}\int_0^x f(t)dt = \lim_{x\to\infty } \frac{f(x)}{2x} $ $\lim_{x\to\infty } \frac{f(x)}{x} =2$
26.08.2019 21:02
dominicleejun wrote: i dont get this qn because $ 1= \lim_{x\to\infty } \frac{1}{x^2}\int_0^x f(t)dt = \lim_{x\to\infty } \frac{f(x)}{2x} $ $\lim_{x\to\infty } \frac{f(x)}{x} =2$ The issue is that $ f $ may not be primitivable. So, you can't use l'Hôpital's rule.
27.08.2019 04:39
f could be $2x + sin(x) + x cos(x)$ $\frac{d}{dx}(x^2 + x sin(x)) = 2 x + sin(x) + x cos(x)$ $ \lim_{x\to\infty } \frac{x^2 + x sin(x)}{x^2} =1. $
27.08.2019 08:47
dominicleejun wrote: f could be $2x + sin(x) + x cos(x)$ $\frac{d}{dx}(x^2 + x sin(x)) = 2 x + sin(x) + x cos(x)$ $ \lim_{x\to\infty } \frac{x^2 + x sin(x)}{x^2} =1. $ It's ok.
27.08.2019 12:50
why isnt it an example for (a)?
27.08.2019 19:49
For part a) one can take as a base function $f(x):=2x$ and then here and there, for example at $x=n, n=1,2,\dots$, we may "add" to $f$ some sharp peaks with area $2^{-n}$ (and heights as we want). These peaks won't affect that given limit, but $f/x$ won't have a limit as $x\to \infty$. For part b), one can argue by contradiction, that's, assume $f$ touches $y=(2+\varepsilon)x$ at points $x_i,i=1,2,\dots, x_i\to \infty$. But since $f$ is increasing, after each "touching" at a point $x_i$ , there is a neighbourhood of $x_i$ where the given integral is disturbed too much. I used to wonder why the Romanians gave those type of problems at a school olympiads. Not that it's a difficult problem, but to grasp the philosophy of it needs some experience.
27.08.2019 23:09
dgrovez, that's because it's ingrained in the educational system the mentality that you have to learn much and understand fast (in all subjects, not just math), with an exaggerated emphasis on memory. In a way I like it, because I was kind of forced by competent people around me to learn hardcore analysis and linear algebra at 10-11th grade. But the sacrifice is the lack of domination of discrete maths (i.e., minimum knowledge, maximum difficulty) like is being done in Russia, per example, which is a shame, since studying nontrivial elementary problems is the best way to get inspiration for superior maths. Per example, I've got this result while solving some elementary problem, which I want to generalize and apply it to analysis. But what I'll get in the end is not as spectacular as the first idea.
29.08.2019 11:05
@CatalinBordea: Oh, I see. I could calculate some easy triple integrals around 8-th degree . To me then, the derivative was $\frac{\Delta f}{\Delta x}$ and I imagined $\Delta x$ to be an infinitesimal small. The definite integral was just a Riemann sum with very small $\Delta x_i$ or some area (volume) interpretation. You see, it wasn't on the rigorous ground. I was just playing, not very interesting to make things coherent, it came later. The good thing was no one pushed me and insisted on this. I don't like this problem especially, because it's a kind of routine thing for undergrads, as opposed to some other good high school Olympiad problems.