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Superior limit of sin(n), and sin(n) always greater than epsilon>0.

1-1 For any \(\epsilon >0\), Prove that there exists \(\sin { (n) } <\epsilon\) where n is natural number. (If \(\epsilon >1\), this is trivial.)

1-2 Prove that there are infinitely many natural number n that satisfies 1-1

2 Using above, prove that \( \limsup _{ n\rightarrow \infty }{\sin { (n) } } =1\).

Note by Min-woo Lee
2 years, 7 months ago

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