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What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity?
In order to solve the following limit $\lim_{n\to\infty} n\sin2\pi n!e$ . This question is very likely to have been asked.
I remember this question and the answer is like $2\pi$ or something .
I also do remember approximating $n!e$ but somehow i don't remember and can't figure out right now .