calculus - Proof of $limlimits_{x to infty}xsin x$ divergences to not infinity($infty, -infty$)

$\lim\limits_{x \to \infty}x\sin x$ divergences to not infinity($\infty, -\infty$)

1. Prove $\lim\limits_{x \to \infty}x\sin x \neq \infty, -\infty$
For $\lim\limits_{x \to \infty}x\sin x = \infty$, $N$ must exist in below:
For all $M>0$, there exists $N$ such that
$x>N \implies x\sin x > M$
But $xsin x$ has lower bound and upper bound for every $x$ ($-x\le x\sin x\le x (x>0)$)
So $x>N \implies x\sin x > M$ can't be satisfied for every $M$ and $N$ does not exist
Therefore $\lim\limits_{x \to \infty}x\sin x \neq \infty$
We can prove $\lim\limits_{x \to \infty}x\sin x \neq -\infty$ in similar way

2. Prove $\lim\limits_{x \to \infty}x\sin x \neq L$ for any constant $L$
For $\lim\limits_{x \to \infty}x\sin x = L$, $N$ and $L$ must exist in below:
For all $\epsilon>0$, there exists $N$ and $L$ such that
$x>N \implies |x\sin x-L| < \epsilon$
$|xsin x-L|$ has following possibilities
1) $|-x-L| \le |x\sin x-L| \le |x-L|$
2) $|x-L| \le |x\sin x-L| \le |-x-L|$
3) $|x\sin x-L| \le |-x-L|, |x\sin x-L| \le |x-L|$
So $|x\sin x-L| < \epsilon$ can't be satisfied for every $\epsilon$,$L$ and $N,L$ does not exists
Therefore $\lim\limits_{x \to \infty}x\sin x \neq L$ for every $L$

Is this proof correct?