prove if f is continuous and $\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$ then $\int\limits _{1}^{\infty}f\left(x\right)\sin x\,dx$ diverges

Without loss of genrality we may suppose $L >0$. If the intergal converges then $\int_a^{b} f(x)\sin x dx$ tends to $0$ as $b >a \to \infty$. Let us pove that this is not true. Consider $\int_{2n\pi}^{2n\pi +\pi} f(x)\sin x dx$. Note that $\sin x$ is positve in this interval Also, $f(x) >\frac L 2$ in this interval if $n$ is large enough. This gives $\int_{2n\pi}^{2n\pi +\pi} f(x)\sin x dx >\frac L 2 \int_{2n\pi}^{2n\pi +\pi} \sin x dx=L$. This is a contradiction.