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$\lim\limits_{x \to \infty}x\sin x$ 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?

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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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  • $\begingroup$ $-1 \le sinx \le 1$ -> $-x \le xsinx \le x (x > 0)$ $\endgroup$ – firia2000 Jun 13 at 9:18
  • $\begingroup$ $a < x < b$ 1) $a \ge 0, x \ge 0, b \ge 0$ 2) $a < 0, x \ge 0, b \ge 0$ 3) $a < 0, x < 0, b < 0$ 4) $a < 0, x < 0, b \ge 0$ $\endgroup$ – firia2000 Jun 13 at 9:22
  • $\begingroup$ $a \ge 0, x \ge 0, b \ge 0$ -> $|a| < |x| < |b|$ $\endgroup$ – firia2000 Jun 13 at 9:27
  • $\begingroup$ $a < 0, x \ge 0, b \ge 0$ -> $|x| < |a|, |x| < |b|$ or $|a| < |x| < |b|$ $\endgroup$ – firia2000 Jun 13 at 9:28
  • $\begingroup$ $a < 0, x < 0, b < 0$ -> $|b| < |x| < |a|$ I missed this so I'll add this $\endgroup$ – firia2000 Jun 13 at 9:29

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