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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.

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    $\begingroup$ Hint: Sandwich Theorem. $\endgroup$ – Jerry Jul 16 '18 at 6:48
  • $\begingroup$ Are you sure it is $x\to\infty$ and not $x\to0$? $\endgroup$ – egreg Jul 16 '18 at 7:31
  • $\begingroup$ 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? $\endgroup$ – rodorgas Jul 16 '18 at 11:24
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    $\begingroup$ @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. $\endgroup$ – egreg Jul 16 '18 at 11:26
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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.

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  • $\begingroup$ What's the criteria for choosing the leftmost and rightmost expressions? Is it removing the "problematic" part, this is, the $\sin$? Thanks! $\endgroup$ – rodorgas Jul 16 '18 at 7:02
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    $\begingroup$ @rodorgas Essentially that, yes. You remove something that is bounded and oscillating and replace it with its bounds. $\endgroup$ – Parcly Taxel Jul 16 '18 at 7:03

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