How I prove that $\lim_{x\to\infty} x \sin x$ does not exist?

Question

How I prove that this limit does not exist? $$\lim_{x\to\infty} x \sin x$$

I can't find two series that disprove this limit. What happens if I use a $2\pi n$ series? $\infty \times 0$ is indefinite.

Anyone have an idea?

Answer 1

For each $n\in\Bbb N$, $\pi n\sin(\pi n)=0$, and therefore $\lim_{n\to\infty}\pi n\sin(\pi n)=0$.

On the other hand,$$\lim_{n\to\infty}\left(\frac\pi2+2\pi n\right)\overbrace{\sin\left(\frac\pi2+2\pi n\right)}^{\phantom0=1}=\infty.$$

Answer 2

Notice, $\sin x$ is a bounded function which fluctuates between $-1$ & $+1$ i.e. $-1\le\sin x\le1\ \forall \ \ x\in \mathbb R$ $$\therefore \lim_{x\to \infty}x\sin(x)=\lim_{x\to \infty}x\cdot \lim_{x\to \infty}\sin(x)=\underbrace{\lim_{x\to \infty}x}_{\to \infty}\cdot \underbrace{\lim_{x\to \infty}\sin(x)}_{[-1,1]}= \infty$$