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

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.

• Hint: Sandwich Theorem. – Jerry Jul 16 '18 at 6:48
• Are you sure it is $x\to\infty$ and not $x\to0$? – egreg Jul 16 '18 at 7:31
• Yes @egreg it's $x \to \infty$. I'm curious, are you asking this because if it was $x\to 0$, the substitution would work? – rodorgas Jul 16 '18 at 11:24
• @rodorgas Essentially yes. You could write the limit as $\displaystyle\lim_{x\to0}\frac{\sin\sin4x}{\sin4x}\frac{\sin4x}{4x}\frac{4}{5}$. Each fraction can be computed with a substitution: the key is that, for instance, $\sin4x$ is invertible in a neighborhood of $0$, so for the first you can substitute $u=\sin4x$. This is not possible for the limit at $\infty$, because the sine is not invertible over unbounded intervals. – egreg Jul 16 '18 at 11:26

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.
• What's the criteria for choosing the leftmost and rightmost expressions? Is it removing the "problematic" part, this is, the $\sin$? Thanks! – rodorgas Jul 16 '18 at 7:02