# Prove that $F=\left(-x^2+2\right)\cdot \cos\left(x\right)+2x\sin\left(x\right)$ don't have limit

We have $f\colon\mathbb{R}\rightarrow \mathbb{R}$, $f\left(x\right)=x^2\cdot \sin\left(x\right)$ and $F$ its primitive. We have to prove that $F$ doesn't have a limit at $\infty$.

What I can say is just that primitive is equal with: $F(x)=\left(-x^2+2\right)\cdot \cos\left(x\right)+2x\sin\left(x\right)$.

In my book the author assumed that $\left(x′_n\right)_{_{n\ge 1}}=2n\pi$ which involving $\lim _{n\to \infty }F\left(x′_n\right)=-\infty$ and $\left(x′′_n\right)_{_{n\ge 1}}=2n\pi +\frac{\pi }{2}$, which involving $\lim _{n\to \infty }F\left(x′′_n\right)=\infty$

My question: Where the author concluded that $\left(x′_n\right)_{_{n\ge 1}}=2n\pi$ and $\left(x′′_n\right)_{_{n\ge 1}}=2n\pi +\frac{\pi }{2}$ and why he proposed it? What is the reason for which did it? What exactly mean to show that haven't limit?

Is my first time when I meet this type of exercise and I want to explain all steps and the purpose. If you have another method is welcome, but explain all steps which led you to did it.

• If $\lim_{x\to\infty}f(x)$ exists and is equal to $L$ then for every sequence $x_n\to\infty$ we must have $\lim_{n\to\infty}f(x_n)=L$. Therefore a useful technique to show non-existence of limits is to find two sequences $x_n\to\infty$ and $y_n\to\infty$ such that $f(x_n)$ and $f(y_n)$ tend to different limits. The idea of using the specific numbers chosen in this proof come from using values in which the trigonometric functions are easy to evaluate, and taking into account their periodicity. Many other choices would have worked as well. – Alamos Apr 25 '15 at 19:34

Even if we do not determine the primitive explicitly, we can show that it cannot have a limit at $\infty$. If the limit existed, then for each $\epsilon>0$ there woiuld exist an $x_0$ such that for all $x,y>x_0$ we'd have $|F(x)-F(y)|<\epsilon$. If we let $x=2n\pi+\frac\pi3$ (with $n$ to be specified in a minute) and $y=x+\frac\pi3$, we note that $\sin t>\frac12$ for $x<t<y$, hecne $f(t)>2n^2\pi^2$ for $x<t<y$, so that certainly $F(y)-F(x)>\frac23n^2\pi^3$. Thus if we pick $n$ so large that both $2n\pi>x_0$ and $\frac23n^2\pi^3>\epsilon$, we arrive at the contradiction we desire and can conclude that the limit does not exist.

• The most important thing is to use their periodicity, didn't it? – Andrei Mihai Apr 25 '15 at 19:45
• @AndreiMihai Not even that (alos, $f$ and $F$ are not periodic strictly speaking). It is enough to have long enough intevals at hand where we can nicely estimate from below. Starting with the even more nonperiodic $f(x)=x^2\sin(x^2)$ the above method still leads to success. – Hagen von Eitzen Apr 25 '15 at 20:23

Let's start with a simpler one because the one you give is more complicated than it needs to be to illustrate the idea. So instead let's just look at

$f(x) = x \sin{(x)}$.

There are many ways to state the condition for a function to have limit as its argument goes to infinity. One way is something like

If, for any x, there exists M such that |f(y) - L| < M for all y>x then f(x) has a limit as x -> infinity and the limit is L.

So, looking at the simple example (graph it you have a hard time visualizing the function) do you see that it fails to meet the criterion I've just given? Notice, $x \sin{(x)}$ oscillates about 0 and its oscillations get bigger and bigger as x increases. So, choosing an arbitrary x there is no value, M, so that $y \sin{(y)}$ is smaller than M for all y>x. Now apply this reasoning to the function that you are looking at.