If $\lim \limits_{x \to \infty}f(x) = L \neq 0$ must it be that $\lim_{x \to \infty} f(x) \sin x$ does not exist?

I got this question:

Let $f\colon \mathbb{R} \to \mathbb{R}$ be a function that satisfies $\lim \limits_{x \to \infty}f(x) = L \neq 0$, Must it be that $\lim_{x \to \infty} f(x) sin x$ does not exist?

I tried to prove it and to find a counter example but I failed so far.

If $\lim_{x\to\infty} f(x)=L$, with $L\ne 0$, then
$$\lim_{n\to\infty} f(2n\pi x)=L$$ and $$\lim_{n\to\infty} f(2n\pi x+\pi/2)=L.$$ But $$f(2n\pi x)\sin(2n\pi )=0,$$ and hence $\lim_{n\to\infty} f(2n\pi x)\sin(2\pi n)=0$, while $$f(2n\pi x+\pi/2)\sin(2n\pi+\pi/2)=f(2n\pi x+\pi/2)\to L,$$ as $n\to\infty$, and hence the limit $$\lim_{x\to\infty} f(x)\sin x$$ does not exist.
Hint: consider the sequence $x_n=\pi/2+n\pi$