(Glancing past the fact that one can do this by hand in a second...) Are there any extra conditions I can give Mathematica to make it evaluate limits like this? I'm not sure why it's getting choked up. enter image description here


k = Sqrt[(2 m \[CapitalEpsilon])/h^2];
\[Mu] = Sqrt[(2 m)/h^2 (\[CapitalEpsilon] + V)];
T = 1/(1 + (k^2 - \[Mu]^2)^2 Sin[2 \[Mu] a]^2/(4 k^2 \[Mu]^2));
Limit[T, \[CapitalEpsilon] -> \[Infinity]]

Assuming your parameters are real, you can replace Sin by its range:

Limit[T /. Sin[_] :> Interval[{-1, 1}], Ε -> ∞]
(*  1  *)
  • $\begingroup$ +1 but E should be replaced by another symbol throughout. $\endgroup$ – Bob Hanlon Aug 9 '15 at 21:07
  • $\begingroup$ @BobHanlon That's a \[CapitalEpsilon], not an E. Still, I would agree even that is a poor choice (potentially confusing). Limit would complain if it were an E. $\endgroup$ – Michael E2 Aug 9 '15 at 21:09
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    $\begingroup$ My oversight. Sorry. $\endgroup$ – Bob Hanlon Aug 9 '15 at 21:12
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    $\begingroup$ @BobHanlon But that's exactly why it's a poor choice. Still, it's the OP's code.... $\endgroup$ – Michael E2 Aug 9 '15 at 21:13
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    $\begingroup$ @Solarmew Thanks & you're welcome. Yeah, sometimes I get annoyed that N etc. are built-in symbols, esp, since N is such a common variable. (I sometimes use NN in such cases, so I'm less likely to get confused.) $\endgroup$ – Michael E2 Aug 10 '15 at 18:48

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