# Proof of $\lim\limits_{x \to \infty}x\sin x$ 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?

I dont understand how you deduce your inequalities (for example if $$x=n\pi$$) Hint: Consider sequences $$n\pi \sin(n\pi)$$ and $$(\pi/2+2n\pi)\sin(\pi/2+2n\pi)$$.
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