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well if you use the substitution 1/y = x the formula becomes

1/y sin(y) or sin(y)/y. Now near where y = 0 (x->infinity)

sin(y)/y is approximately 1. So does that help? I'm not sure what

the hausdorff distance is. It's been a very long time since I've taken

a math course.

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15y ago

What else can I help you with?