This question already has an answer here:

So I got this assignment. And I was wondering how is it possible to get a limit from a constantly changing formula. $$\lim_{x\to+\infty}\frac{\sin(x)}{x}$$ Can I only look in the domain $]0,2\pi[$?


marked as duplicate by Martin Sleziak, user296602, Dan, Najib Idrissi, jameselmore Jan 20 '16 at 20:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Hint: $|\sin x|\le1$. $\endgroup$ – David Mitra Oct 31 '15 at 12:24
  • 6
    $\begingroup$ @Socre: Your analysis is not well-founded. $\endgroup$ – hardmath Oct 31 '15 at 12:29
  • 1
    $\begingroup$ @Socre: I'm pretty sure that de l'Hospital is not applicable in this case... $\endgroup$ – Fabian Oct 31 '15 at 12:48
  • 2
    $\begingroup$ @Socre L'Hopital's rule only applies when both numerator and denominator go to infinity or both go to 0. That doesn't happen here. Plus, if you just plot this function, it's obvious what happens. $\endgroup$ – Milo Brandt Oct 31 '15 at 13:26
  • 5
    $\begingroup$ math.stackexchange.com/questions/1368086/… $\endgroup$ – Von Neumann Jan 20 '16 at 19:27

A possible approach is to take help of the following:

  1. $$\lim_{x\to+\infty}\frac{1}{x}=0$$

  2. $$-1\le\sin x\le1$$

  • 3
    $\begingroup$ This is wrong because $\sin x$ does not approach "a finite rational number" as $x\to \infty$. $\endgroup$ – hardmath Oct 31 '15 at 12:31

No, you cannot only analyze the problem within $]0, 2\pi[$, although $\sin$ is periodic.

Yes, the limit exists and $=0$. We have $x > 0$ only if $$ \bigg| \frac{\sin x}{x} \bigg| \leq \frac{1}{x}; $$ given any $\varepsilon > 0$, we have $1/x < \varepsilon$ if $x > 1/\varepsilon$; hence we have proved this: for every $\varepsilon > 0$, we have $x > 1/\varepsilon$ only if $$ \bigg| \frac{\sin x}{x} \bigg| < \varepsilon; $$ that is, $$ \lim_{x \to +\infty}\frac{\sin x}{x} = 0. $$


You can't look only in $]0,2\pi]$, as this function is not periodic. The shortest way is to observe $\dfrac{\sin x}x=O\Bigl(\dfrac1x\Bigr)$, which tends to $0$ as $x$ tends to $\infty$.

  • 4
    $\begingroup$ Most beginning calc students don't know big O notation, but they know the Squeeze Theorem... $\endgroup$ – Thomas Andrews Oct 31 '15 at 12:30
  • $\begingroup$ Maybe he/she'll have the curiosity to delve into asymptotic analysis seeing how efficient and short it is on simple cases. I agree on this example, the Squeeze theorem is just as good. $\endgroup$ – Bernard Oct 31 '15 at 12:36

First note that $$-1\leq\sin x\leq 1$$ $$-\frac1{|x|}\leq\frac{\sin x}{|x|}\leq \frac1{|x|}$$ Using the squeeze theorem, we have $$-\lim\limits_{x \to \infty} \frac1{|x|}\leq\lim\limits_{x \to \infty} \frac{\sin x}{|x|}\leq\lim\limits_{x \to \infty} \frac1{|x|}$$ As $x\to\infty$, we have $|x|=x$. So we can drop the absolute value bars $$-\lim\limits_{x \to \infty} \frac1{x}\leq\lim\limits_{x \to \infty} \frac{\sin x}{x}\leq\lim\limits_{x \to \infty} \frac1{x}$$ $$0\leq\lim\limits_{x \to \infty} \frac{\sin x}{x}\leq 0$$ Therefore $$\lim\limits_{x \to \infty} \frac{\sin x}{x} =0$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.