I want to prove that
$$ \nexists \lim_{n \to \infty} \frac{\cosh (ny) \sin (nx)}{n^3} $$
for any $x \notin \pi\mathbb{Z}$, $y \neq 0$.
I think it is obvious, but I don't find a way to prove it rigurously. This is what I've done so far:
Suppose $\lim_{n \to \infty} \frac{\cosh (ny) \sin (nx)}{n^3} = L \in \mathbb{R}$. Now, we have
$$ \frac{\cosh (ny) \sin (nx)}{n^3} = \frac{1}{2} \frac{e^{ny}}{n^3}\sin (nx) + \frac{\sin(nx)}{2n^3 e^{ny}}$$
We can assume wlog that $y>0$. Now, because $(\sin (nx))_n$ is a bounded sequence and $\lim_{n \to \infty} 2n^3e^{ny}=0$ (if $y <0$ we would have taken $\lim_{n \to \infty} 2n^3e^{-ny}=0$), we know that
$$ \lim_{n \to \infty}\frac{\sin(nx)}{2n^3 e^{ny}}=0 $$
So
$$ \lim_{n \to \infty} \frac{e^{ny}}{n^3} \sin (nx) = 2L $$
My intuition now says that the product of a sequence that diverges to infinity with a divergent sequence that is bounded is always divergent. But I've searched a lot and I haven't found any similar result.
The most closely related results I've found use subsequences, but the only case I think it can be applied is to $x= \frac{\pi}{2}$, since $(\sin (\frac{n\pi}{2}))_n=(0,1,0,-1,0,1,\dots)$ so you can take a subsequence that diverges.
Edit:
As someone commented, it can be proven that $\sin (nx)$ cannot tend to zero if $n \notin \pi \mathbb{Z}$ but it remains unclear to me where to find the result that states that if $\lim_{n \to \infty} f(n)= \infty$ and $\lim_{n \to \infty} g(n) \neq 0$ then $\nexists \lim_{n\to \infty} f(n)g(n)$
Since $e^{ny}/n^3\to \infty$ when $y>0$ (for $y<0$ the argument is similar) we only need to show that $sin(nx)\not\to 0$; hot establish this, see why sin(n) is dense in[-1,1] and here the argument is similar.