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Bug introduced in 9.0 and fixed in 10.0


When I evaluate the limit of (1-Cos[x]) Sin[1/x] at x = 0 in Mathematica 9, I obtained Interval[{-2, 2}]. The problem is that this limit is 0. It is also 0 when I compute it by Maple and Mathematica 7. So, I wonder whether I've found a bug in Mathematica 9?

Limit[(1 - Cos[x]) Sin[1/x], x -> 0]
Interval[{2, 2}]
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    $\begingroup$ Are you sure that you didn't write by mistake (1+Cos[x]) Sin[1/x]? $\endgroup$ – Spawn1701D Apr 11 '13 at 1:45
  • $\begingroup$ @Spawn1701D No. Please see the image added in my questions. Thank you for your answer! $\endgroup$ – Z-Y.L Apr 11 '13 at 1:59
  • $\begingroup$ What subversion of 9 you have? Did you run the command after a initializing the kernel? If this is true its a big mess up! It's like saying that $\cos(0)=-1$! $\endgroup$ – Spawn1701D Apr 11 '13 at 2:04
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    $\begingroup$ It's working fine for me on 9.0.1. Make sure x doesn't have a value with x=. first. $\endgroup$ – Michael Hale Apr 11 '13 at 2:08
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    $\begingroup$ So we know it's a 32-bit / 64-bit issue. @Z-Y.L I guess there's not much more to say than that it's a bug ... could you report it to support@wolfram.com? $\endgroup$ – Szabolcs Apr 11 '13 at 13:37
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This bug has been fixed as of Mathematica 10.0.0.

Limit[(1 - Cos[x]) Sin[1/x], x -> 0]

(* 0 *)
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