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Show $\displaystyle\lim_{x\to\infty}x\sin x$ does not exist using Cauchy definition of limit.

Am supposed to negate the defintion: $\exists \epsilon >0 :\forall \delta>0 :(\forall x: 0<|x-a|>\delta \Rightarrow |f(x)-L| > \epsilon) $ ?

But here $a$ is infinity and there is no $L$ so I don't know how to write the inequality.

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  • $\begingroup$ Cauchy has a different definition in the case of infinite limits. en.wikibooks.org/wiki/Calculus/Formal_Definition_of_the_Limit $\endgroup$ – grantfgates May 7 '14 at 14:58
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    $\begingroup$ Use the transformation $u=1/x$, then the limit becomes $\lim_{u\rightarrow 0^+}\frac{\sin 1/u}{u}$ $\endgroup$ – Samrat Mukhopadhyay May 7 '14 at 14:58
  • $\begingroup$ What does $0<|x-a|>\delta$ mean? What could this possibly mean here (where "$a$ is infinity", as you say)? $\endgroup$ – Andrés E. Caicedo May 7 '14 at 15:00
  • $\begingroup$ @AndresCaicedo seeing from the definition for infinite limits, $a$ becomes $0$. What it is ? I'm not really sure, I tried to negate to statement, I think It's similar to "for $n>N\in \mathbb N$" from sequences. $\endgroup$ – GinKin May 7 '14 at 15:05
  • $\begingroup$ @GinKin I think you've made a mistake in you definition, I shoul mention. Pleas check it. $\endgroup$ – gebruiker May 7 '14 at 15:35
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To show that the limit doesn't exist using the formal definition of a limit, I would do something like this:

Suppose $\lim\limits_{x\to\infty}x\sin x=L$. Then we should have: $$\forall\epsilon>0\, \exists N\in\mathbb{N}:x>N\Rightarrow|x\sin x-L|<\epsilon$$ However \begin{equation}|x\sin x-L|\geq|x\sin x|-|L|=|x||\sin x|-|L|\end{equation} Now by choosing $x=\frac\pi2+2Nk\pi$ ($k\in\mathbb{Z}_{\geq0}$) we can make this quantitiy arbitrairilly high, while still having $x>N$. This is a contradiction. Hence the limit doesn't exist.

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