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I'm trying to determine the limit of the sequence of functions

$$f_n(x)=\left(\frac{1}{\pi}\arctan(n x) + 1/2\right)^n. $$

I define

f[x_, n_] := (1/2 + ArcTan[n x]/Pi)^n

And enter

Limit[f[x, n], n -> Infinity]

This gives the answer 0.

If I instead enter

Assuming[x > 0, Limit[f[x, n], n -> Infinity]]

I get the answer $$e^{-\frac{1}{x\pi}}. $$

While

Assuming[x < 0, Limit[f[x, n], n -> Infinity]]

gives the answer 0 again.

Why does the normal Limit give the answer assuming that x<0? Is this a bug, or have I missed something?

Thanks in advance.

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Please only use the tag bugs post-hoc, if it is confirmed by WRI or the community. –  István Zachar Oct 16 '13 at 12:19
1  
I think it's a bug. The result is not always 0 as you check by using 1 for instance in place of x. –  b.gatessucks Oct 16 '13 at 12:51
6  
Probably a bug or at least a limitation in calculus code. Limit will rely on Series and at least for elementary functions that will ignore branch cut issues unless assumptions are provided that give a clear indication of what side of such a cut we are on. –  Daniel Lichtblau Oct 16 '13 at 15:55

1 Answer 1

f[x_, n_] := (1/2 + ArcTan[n x]/Pi)^n;
Limit[f[x, n], n -> Infinity, Assumptions -> x > 0]
E^(-(1/(Pi x)))
share|improve this answer
2  
I think the original poster may have mentioned that.. –  Daniel Lichtblau Sep 12 at 21:09

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