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.
Answer
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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