What is

$$\lim_{x\to\infty} \sin x$$


I always thought it was undefined; however, Wolfram|Alpha says that

$$\lim_{x\to\infty} \sin x = -1 \text{ to } 1$$

Now, what is the correct answer? What does it even mean if the limit is not a single, distinct value, but rather an interval?

Other than that, am I right in assuming that

$$\liminf_{x\to\infty} \sin x = -1$$


$$\limsup_{x\to\infty} \sin x = 1$$ ?

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    $\begingroup$ They mean that the values of $\sin(x)$ for $x\to+\infty$ accumulate to every value in $[-1,1]$. In particular, that implies the $\limsup$, $\liminf$ equations that you wrote, and that in the context of an introduction to calculus, the limit doesn't exist. $\endgroup$ – user561348 May 24 '18 at 12:37
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    $\begingroup$ @arugula so is it in fact undefined? $\endgroup$ – user500664 May 24 '18 at 12:37
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    $\begingroup$ Who knows what Alpha means, but I guess you could interpret it as "for every point $p$ between -1 and 1, there's a sequence tending to infinity, such that $\sin$ of that sequence is constantly $p$". And yes you're right about the sup and inf limits. $\endgroup$ – Steve D May 24 '18 at 12:38
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    $\begingroup$ To find what Wolfram Alpha means may be difficult? But presumably this is the same answer as Mathematica itself, which does have extensive documentation you can look at. $\endgroup$ – GEdgar May 24 '18 at 12:45

Take note that the sine function takes values between $-1$ and $1$. When you are talking about a limit to $\infty$, it's an undetermined number, which is infinitely large. Now, taking into account the periodicity of the sine function, there is no possible way to determine a specific value, as it entirely depends on the nature of the "infinite" number.

More specifically, $-1 \leq \sin(x) \leq 1, \; \; \forall x \in \mathbb R$. This means that for any given $x$ over the real numbers, the sine function is bounded. Thus, all you can say about an undetermined infinite limit (it does not exist talking strictly mathematics), is :

$$-1 \leq \lim_{x \to \infty} \sin(x) \leq 1$$

What you mentioned though is indeed true :

$$\liminf_{x\to\infty} \sin x = -1$$

$$\limsup_{x\to\infty} \sin x = 1$$

  • $\begingroup$ So it can be understood as "the limit doesn't exist, but if it existed it would be between -1 and 1", right? $\endgroup$ – user500664 May 24 '18 at 12:42
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    $\begingroup$ Essentially, yes. The limit does not exist, but it surely is bounded. $\endgroup$ – Rebellos May 24 '18 at 12:43
  • $\begingroup$ alright, thank you. I will accept your answer as soon as it will let me. $\endgroup$ – user500664 May 24 '18 at 12:43

The limit doesn't exist and it can be proved formally by 2 subsequences with different limits, that is

  • $x_n=2n\pi+\frac{\pi}2 \to \infty \implies \sin(x_n)=1$

  • $x_n=2n\pi+\frac{3\pi}2 \to \infty \implies \sin(x_n)=-1$

Yes what is true is that $\liminf=-1$ and $\limsup=1$ indeed

$$-1\le\sin x \le 1$$

and we have found 2 subsequences which tends to those limits.

  • $\begingroup$ Thank you for the proof as well. Your answer is very good, but I promised Rebellos to accept his answer earlier, so I am sorry. Thank you very much though. $\endgroup$ – user500664 May 24 '18 at 12:55
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    $\begingroup$ @ThomasFlinkow You are welcome! The choice of the answer to be accepted is absolutely up to you! Bye $\endgroup$ – user May 24 '18 at 13:05

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