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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 –  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

2 Answers 2

up vote 16 down vote accepted

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 ; 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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