We have the following function:
$$U_n = \sin \dfrac{1}{3} n \pi$$
What is the limit of this function as n approaches infinity?
I first tried to use my calculator as help, for n I chose some arbitrary large numbers, such as 100 and 1000. Then I I just took $n = 10^{50}$ and it gave me an error.
So the correct answer is it doesn't have one, but why? Why does this function have a solution for $n = 10^2, 10^3$ and not for bigger numbers such as $n=10^{50}$?
Answer
As David hinted in his comment, try using the fact that
$$
\sin(x + 2\pi k) = \sin(x)
$$
whenever $k$ is an integer. Or, if you just write out the first few values of $U_n$ (compute these using the unit circle) and you should notice a pattern.
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