# Why is the output of my limit expression an interval?

Why is the output of the limit below an interval? It should be precisely $1$.

Limit[(2/Pi) ((2 n + 2)!!/(2 n + 1)!!) Integrate[(1 - x^2)^(n + 1/2), {x, 0, 1}],
n -> Infinity]

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You have made an assumption without telling Mathematica about it. Namely, that $n$ is an integer. You can add this assumption like so:
Limit[(2 (2n + 2)!! Integrate[(1 - x^2)^(n + 1/2),

Without this assumption it appears that the correct interval of fluctuation is 2/Pi to 1. Limit[] is unable to completely sort out the dependencies of the various fluctuating components. I think it comes reasonably close though; if you blindly plug in for n an interval of length 2Pi, for large n, the interval range that results is far bigger. –  Daniel Lichtblau Feb 28 '13 at 15:06
@Daniel Using FunctionExpand on the expression first gives {2/Pi, 1}. Does FunctionExpand do any transformations that are not generally valid? –  Szabolcs Mar 1 '13 at 0:35
@Danel I used it as Limit[FunctionExpand[(2/ Pi) ((2 n + 2)!!/(2 n + 1)!!) Integrate[(1 - x^2)^(n + 1/2), {x, 0, 1}]], n -> \[Infinity]] I was just wondering why Limit doesn't try to use it itself. But how these symbolic functions can choose a strategy to calculate something is a big mystery to me :) It must be quite complicated and difficult to get right. –  Szabolcs Mar 1 '13 at 15:51