The question is how to prove that $\lim_{n \rightarrow \infty} \left(\dfrac{\sin{n}+\sin^2n^2+\sin^3n^3} {n^3} \right) = 0$. I know $|\sin(n)| ≤ 1$,but I don't know the range of $\sin^2n^2$, $\sin^3n^3$ and the sum of these three together, could any one help me to prove it? Thanks!
How to prove that $\lim ((\sin(n)+\sin^2(n^2)+\sin^3(n^3)) / n^3 ) = 0$.
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calculus
analysis
limits
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0Detail : you know that $|\sin(n)|\leq 1$ – 2017-01-28
1 Answers
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hint: $|\sin^k x| \le 1, k \ge 1\implies -\dfrac{3}{n^3} \le Q(n)\le \dfrac{3}{n^3}$
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0But I am not sure for sin² is it always positive? So the sum of them cannot be -3? And is the power on n makes any effect? – 2017-01-28
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0@BJin Why should that matter? The sine of *anything* is $\le 1$ in absolute value, and such an expression raised to *any* natural power is still $\le 1$ in absolute value. – 2017-01-28