4
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Why is this limit not working?

Limit[I*Sin[alpha^0.5*Betta]*Sin[Lc*Betta*alpha^0.5]/(alpha^0.5*Betta),alpha -> 0]

If I remove the division by alpha it works fine, but it doesn't if I divide the two Sines by alpha

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5
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    $\begingroup$ try (1/2) or Sqrt instead of 0.5 $\endgroup$ – george2079 Mar 27 '17 at 21:38
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    $\begingroup$ @george2079 Actually, if I change only one of the two 0.5 into 1/2 (leaving the other one 0.5) I get a result. Weird? The weirdest thing is that I dont need to change both 0.5 in 1/2. $\endgroup$ – Andrea G Mar 27 '17 at 22:00
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    $\begingroup$ Btw, that greek letter you are referring to is not called Betta. Actually, it is called Bettina and it looks like this :) $\endgroup$ – halirutan Mar 28 '17 at 4:07
  • $\begingroup$ @george2079 Do you know why it fails with 0.5? $\endgroup$ – anderstood Mar 28 '17 at 14:44
  • $\begingroup$ Its typical that using inexact real numbers (.5) for things that should be mathematically exact cause such problems. Its not always predictable of course, Limit[Sin[x^.5]/x^.5,x->0] works when you might expect it should not. $\endgroup$ – george2079 Mar 28 '17 at 15:08
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Rewriting this way we can reach for some result:

Limit[I*Sin[Sqrt[α]*β]*(Sin[Lc*β*Sqrt[α]]/(Sqrt[α]*β)),α->0]

0

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