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I tried to calculate the following limit $$\lim_{n\to\infty}n\sin(2\pi en!)$$ Mathematica gets

In[1]:= Limit[n*Sin[2 Pi*n!*Exp[1]], {n -> Infinity}]


Out[1]= {Interval[{-\[Infinity], \[Infinity]}]

But by the following link the answer is $2\pi$. I'm confused.

http://math.stackexchange.com/questions/76097/what-is-the-limit-of-n-sin-2-pi-cdot-e-cdot-n-as-n-goes-to-infinity

I think the limit is zero.

ListPlot[Table[n Sin[2 Pi E Factorial[n]], {n, 0, 300}]]
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Plot[n Sin[2 Pi E Factorial[n]], {n, 0, 15}] –  BoLe Mar 13 at 9:53
1  
Limit give the limit of the expression as a real function (its divergent, thus the [-inf;inf] ), and not the limit of a sequence. See forums.wolfram.com/mathgroup/archive/2000/Sep/msg00318.html and forums.wolfram.com/mathgroup/archive/1999/Jul/msg00375.html for some strategies to evaluate limits of sequence (spoiler : they don't work here, you have to do the math) –  William Briand Mar 13 at 10:03
    
@WilliamBriand I think the limit is zero. ListPlot[Table[n Sin[2 Pi E Factorial[n]], {n, 0, 300}]] –  alex Mar 13 at 12:50
1  
Try that plot using N[n*Sin[2 Pi E n!],1000]. –  Daniel Lichtblau Mar 13 at 13:28

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