Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon). Question:

Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon). 

What I did:
I've been struggling with this one for a long time. Really tried digging up the net for explanations, but couldn't find a really good source to explain how this proof should be done(some said proof by contradiction like I did, and some say without contradiction). I got too confused by now, I also read this Proving limit doesn&#39;t exist using the $\epsilon$-$\delta$ definition
which gave some guidance but I thought I'd use the floor function to make sure M is natural, otherwise the idea of using these values wouldn't have worked.
Anyway I'd really love someone to verify/correct this solution before I hand it in.
Thanks a lot.


We assume there exists a limit, therefore: 
$\forall \varepsilon&gt;0 \exists M&gt;0: x&gt;M \Rightarrow |f(x)-L|&lt;\varepsilon$
and specifically, if we pick some $x_1,x_2&gt;M$: $|f(x_1)-L|&lt;\frac \varepsilon2, 
|f(x_2)-L|&lt;\frac \varepsilon2$
Also, this applies for all $\varepsilon$, and specifically for  $\varepsilon=0.5$
(1) Therefore using the triangle inequality: $|f(x_1)-f(x_2)| = |f(x_1)-L+L-f(x_2)| \le |f(x_1)-L |  + |f(x_2)-L| \le \frac \varepsilon2 +  \frac \varepsilon2 = \varepsilon$
This applies for all $x&gt;M$ so specifically for: $x_1=\frac {\pi \lfloor M+1 \rfloor}{2} &gt; M , x_2=\pi \lfloor M+1 \rfloor$
We put these values in (1): $|f(x_1)-f(x_2)|=|\frac {\pi \lfloor M+1 \rfloor}{2}\sin\frac {\pi \lfloor M+1 \rfloor}{2}-\pi \lfloor M+1 \rfloor \sin (\pi \lfloor M+1 \rfloor)|= \begin{cases} 1 &amp; \lfloor M+1 \rfloor   odd  \\ 0 &amp;\lfloor M+1 \rfloor  even\end {cases} $
So we have found an $\varepsilon $ that doesn't always satisfy the requirements of the limit definition so this is a contradiction. Therefore the limit doesn't exist.
Solution by means of sequences:

$$x_n:=\pi n\implies x_n\sin x_n=0\xrightarrow[n\to\infty]{}0$$

$$y_n:=\frac{(4n-3)\pi}2\implies y_n\sin y_n=\frac{(4n-3)\pi}2\xrightarrow[n\to\infty]{}\infty$$
you can always choose $x=k\pi$, then you can show by the $\epsilon-\delta$ language that if the limit exists, it must be $0$. But then, you choose $x=2k\pi+\frac\pi2$, which shows that the limit might not be $0$.
Ok, I have used the uniqueness of limits in $R$, but somehow you can always choose some special $x$, which makes $sinx$ become a constant to solve this question.