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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.

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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.

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Hint: consider the sequence $x_n=\pi/2+n\pi$

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