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I want to find out $\displaystyle\lim_{x \to +\infty} x \sin x$. Now, this doesn't exist, but I'm not sure how to transform the definition of limit to something that lets me prove that the limit doesn't exist. This is the definition I use, for the record:

We say that $\displaystyle\lim_{x \to +\infty} f(x) = l$ if $\forall \epsilon > 0, \exists M > 0$ such that $x > M \implies |f(x)-l| < \epsilon$.

This isn't exactly $\epsilon-\delta$, it's more like $\epsilon - M$, but it's the same idea. My problem is: how to use this to prove that the limit doesn't exist? I know that I would have to begin like this:

We say that $\displaystyle\lim_{x \to +\infty} f(x)$ doesn't exist if $\exists \epsilon>0$ such that $\forall M > 0$ . . .

And I don't know how to continue.

Edit: I want to clarify something: while I am indeed trying to prove the nonexistence of $\displaystyle\lim_{x \to +\infty} x \sin x$, the point of this question was to be able to use the definition to prove the nonexistence of any limit, not just this one.

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  • $\begingroup$ You can start by noting that $-x\leq x \cdot \sin x\leq x$ $\endgroup$
    – Pedro
    Commented Apr 14, 2012 at 21:49
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    $\begingroup$ $\lim\limits_{x\rightarrow\infty} f(x)\ne L$ would mean that there is an $\epsilon>0$ such that for any $M>0$, there is an $x>M$ so that $|f(x)-L|\ge \epsilon$. $\endgroup$ Commented Apr 14, 2012 at 21:51
  • $\begingroup$ @DavidMitra: Please post that as an answer, it's pretty much what I'm looking for. $\endgroup$
    – Javier
    Commented Apr 14, 2012 at 21:56

3 Answers 3

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$\lim\limits_{x\rightarrow\infty} f(x)\ne L$ would mean that there is an $\epsilon>0$ such that for any $M>0$, there is an $x>M$ so that $|f(x)-L|\ge \epsilon$.

To use the above to show that $\lim\limits_{x\rightarrow\infty} f(x)$ does not exist, you would have to show that $\lim\limits_{x\rightarrow\infty} f(x)\ne L$ for any number $L$.

For your purposes, with $f(x)=x\sin x$, let $L$ be any number. We will show that $\lim\limits_{x\rightarrow\infty} f(x)\ne L$. Towards this end, take $\epsilon=1$. Now fix a value of $M$. Using Alex's answer, you can find an $x>M$ so that $|f(x)-L|\ge1$.

Thus $\lim\limits_{x\rightarrow\infty} f(x)$ does not exist.

(The limit might be infinite (it isn't, see Alex's answer again); but this is another matter...)

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Just notice that you can make $\pi n$ and $\frac{\pi n}{2}$ arbitrarily large yet $\displaystyle \pi n \sin\pi n=0$ and $\displaystyle \frac{\pi n}{2}\sin \left(\frac{\pi n}{2} \right)=\frac{\pi n}{2}$.

EDIT: To see why this solves your question suppose that the limit exists and equals $L$. Then, there exists $M$ such that $x>m$ implies that $|x\sin(x)-L|<.5$ and so evidently if $x,y>M$ then $|x\sin(x)-y\sin(y)|<1$. See how this stands up with $x=\pi M$ and $y=\frac{\pi M}{2}$.

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  • $\begingroup$ Please see my edit, it seems my question wasn't too clear. $\endgroup$
    – Javier
    Commented Apr 14, 2012 at 21:57
  • $\begingroup$ See my edit for why it solves it. $\endgroup$ Commented Apr 14, 2012 at 22:03
  • $\begingroup$ What about the case where $L=\infty$? $\endgroup$
    – AnnieOK
    Commented May 2, 2014 at 21:12
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You can define two different sequences that diverge to infinity and show that $f(x_n)$ and $f(y_n)$ have different limits. Work with properties of $sin$ here.

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