How to evaluate $\lim_{x\to+\infty}\frac{\sin\sin4x}{5x}$?

Question

I'd like to learn how to evaluate this limit: $$\lim_{x\to+\infty}\frac{\sin\sin4x}{5x}$$ I tried to substitute with a new variable:

Let $u=\sin4x$. Then as $x\to+\infty$, $u\to\,??$

Since $\sin x$ won't stop variating on $+\infty$, I don't know how to evaluate this.

Answer 1

Since $$-\frac1{5x}\le\frac{\sin\sin4x}{5x}\le\frac1{5x}$$ where both leftmost and rightmost expressions tend to 0 as $x\to\infty$, the squeeze theorem gives 0 for the original limit.