>>272457I know that, the problem is that with the n^3, it hard to prove it with the definiton.
What about:
Let's suppose that Lim Sin(n^3) = L
we know that |Sin(n^3)| =< 1
so for all n, we have -1 =< L \=< 1 so with Arcsin/Sin^-1 being continuous, we have
\lim_{n \to \infty} n^3 = \lim_{n \to \infty} \sin^{-1}(\sin(n^3)) = \sin^{-1}(L) =/= +\infty
Is that good?
