The Question
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> let $f:\left[1,\infty\right)\to\mathbb{R}$ be a continuous function with $\lim\limits \_{x\to\infty}f\left(x\right)=L\neq0$. then $\int\limits \_{1}^{\infty}f\left(x\right)\sin x\,dx$ diverges.
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I came across this question while preparing to my exam in calculus. I wasn't able to solve it: couldn't prove it nor give a counterexample.

What have I tried?
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I tried using tests such as the comparsion test, but I realised $f\left(x\right)\sin x$ isn't positive. Tried looking for counter examples but failed.

Here is my best attempt so far:
since $L\neq0$ we can assume without loss of generality $L>0$, therefore $f>0$ almost always.
for some $\mathbb{N}\ni k,\, M=2k$ we have $$\int\limits \_{M\pi}^{k\pi}f\left(x\right)\sin x\,dx=\sum\_{i=M}^{k}\left(-1\right)^{i}\int\limits \_{0}^{\pi}f\left(x\right)\sin x=Q\_{k}$$ and since $$2\left(L+\epsilon\right)\leq\int\limits \_{0}^{\pi}f\left(x\right)\sin x\leq2\left(L+\epsilon\right)$$ for some $\epsilon$, we can write $$\sum\_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i+1}\epsilon\right)\leq Q\_{k}\leq\sum\_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i}\epsilon\right)$$

But here I got stuck, I'm not sure how to continue and I'm not sure if this is the right direction.