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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$\begingroup$If $\sin x$ had a 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 $1$.
$\endgroup$ 8 $\begingroup$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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