# Limit approaching infinity of sine function

I'd like to ask a question which I have been reflecting on for some time now. What is the limit of: $f(x) = \sin(x)$ as $x$ tends to infinity?

As we know, the function has a definite value for each multiple of a value included between $0$ and $2\pi$, but, how can we know which value it will have at infinity?

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Sine isn't defined at infinity, and the limit certainly does not exist. See en.wikipedia.org/wiki/Limit_(mathematics) –  anon Jul 19 '11 at 10:06
Thanks.. Why is it not defined at infinity? –  max0005 Jul 19 '11 at 10:07
Briefly put: for the same reason it isn't defined for apples or other assorted fruits. You could arbitrarily assign $\sin(\text{apples})=1$, but that's meaningless. –  anon Jul 19 '11 at 10:53

If $\sin x$ would have limit $L$ for $x\to\infty$, then for every sequence $(x_n)$ such that $x_n\to\infty$ we would have $$\lim\limits_{n\to\infty} \sin x_n=L.$$ In particular, this limit would exist and would have the same value for every choice of such sequence $(x_n)$. (See e.g. here http://en.wikipedia.org/wiki/Limit_of_a_function#Sequential_limits ; but this theorem was probably mentioned in your lecture/textbook.)

If you choose $x_n= 2n\pi$, then this limit is equal $0$.

If you choose $x_n=\frac\pi2+2n\pi$, then this limit is equal to $1$.

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But aren't $2n\pi$ and $\pi/2+2n\pi$ equal when $n=\inf$? –  max0005 Jul 19 '11 at 10:36
@max I am not sure I understand you question completely, but the point was that both $2n\pi$ and $\frac\pi2+2n\pi$ tend to $+\infty$, so in a sense you can say that "they are equal in infinity". (Which shows that $\sin x$ cannot be "continuously extended" there.) –  Martin Sleziak Jul 19 '11 at 10:42
Well, if both of them are equal at infinity, that would mean that two different non-supplementary values had the same sine value, which of course wouldn't make sense. I was asking myself though if any sort of formal definition existed. –  max0005 Jul 19 '11 at 13:03
@max: Formal definition of limit makes no mention of infinity. If limit of $f(x)$ as $x \to \infty$ is $L$, that simply means that given any $\epsilon > 0$, there exists $\delta>0$ such that $x > \delta \Rightarrow |f(x)-L| < \epsilon$. –  user5501 Jul 19 '11 at 13:28
@max0005: At this stage of the game, you must repeat to yourself "infinity is not a number", "infinity is not a number", "infinity is not a number", $\dots$. Thinking of $\infty$ as a number just invites confusion. Later, in very carefully defined contexts, it may be useful to assign meaning to the symbol $\infty$. –  André Nicolas Jul 19 '11 at 14:25

The set of limit points of $\sin x$ as $x \to \infty$ is $[-1,1]$. In particular, it is not a single point, and thus the limit doesn't exist.

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For all x with |sin(x)| < 1 there exists an y with y > x and |sin(y)| > |sin(x)|. So sin(x) does not converge. It's approaching 0, then becoming more distant and so on.

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Your first sentence, in itself, does not rule out convergence. $x/(1+x)$ has the same property but does converge as $x \to \infty$. For the third sentence, $\frac{\sin x}{x}$ also in some sense "approaches zero and then becomes more distant" but does converge. One needs to be more careful. –  Nate Eldredge Jul 19 '11 at 14:02
@Nate Eldregde: Yes, you're right. Thank you for the explanation. I think I've tried to explain why the limit of $sin(x)$ is not zero. I really have to be more careful. –  Robert Jul 19 '11 at 15:56

limit x tends to infinity then

sin(limit n tends to infinity 2*n*pie) = sin(lim n tends to infinity 2*n*pie + limit h tends to zero h) = limit h tends to zero sin(h) = 0

sin(infinity)=0

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The computation doesn’t make sense, and $\infty$ isn’t in the domain of the sine function. –  Brian M. Scott Mar 28 '13 at 6:33

All you can say is that it oscillates between -1≤ limx→∞ sin(x) ≤1

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jay: the reason this is being downvoted is that $\lim_x \sin(x)$ doesn't exist, so it the inequality you've written down doesn't make sense. –  Gerben Jul 19 '11 at 16:18