# Showing a limit does not exist using Cauchy's $\epsilon, \delta$ limit definition

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.

• Cauchy has a different definition in the case of infinite limits. en.wikibooks.org/wiki/Calculus/Formal_Definition_of_the_Limit – grantfgates May 7 '14 at 14:58
• Use the transformation $u=1/x$, then the limit becomes $\lim_{u\rightarrow 0^+}\frac{\sin 1/u}{u}$ – Samrat Mukhopadhyay May 7 '14 at 14:58
• What does $0<|x-a|>\delta$ mean? What could this possibly mean here (where "$a$ is infinity", as you say)? – Andrés E. Caicedo May 7 '14 at 15:00
• @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. – GinKin May 7 '14 at 15:05
• @GinKin I think you've made a mistake in you definition, I shoul mention. Pleas check it. – gebruiker May 7 '14 at 15:35

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 $$|x\sin x-L|\geq|x\sin x|-|L|=|x||\sin x|-|L|$$ 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.