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Upright [...] in italics quotation



Weird result in complex limit


Limit of integral gives incorrect outputIntegrate returns unexpected resultLimit problem calculating directional derivativeWhy won't Limit evaluate, and what can be done about itLimit of an inverse functionDoes Mathematica implement Risch algorithm? If it does, in which cases?Limit problem no longer works in Mathematica 11.1.0Evaluating integral seems incorrectReal integral giving complex resultHow to apply NIntegrate three times













1












$begingroup$


I am trying to evaluate a limit:



gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]


I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):



$qquad frac s2 sqrtfrac ue$



But for some reason, when using Limit, I get



DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]


So my questions are:



What is going here?

What issues should I be aware of when using Limit?










share|improve this question









New contributor




Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$
















    1












    $begingroup$


    I am trying to evaluate a limit:



    gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
    Limit[Re[gamma[x]], x -> DirectedInfinity[1]]


    I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):



    $qquad frac s2 sqrtfrac ue$



    But for some reason, when using Limit, I get



    DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]


    So my questions are:



    What is going here?

    What issues should I be aware of when using Limit?










    share|improve this question









    New contributor




    Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      1












      1








      1





      $begingroup$


      I am trying to evaluate a limit:



      gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
      Limit[Re[gamma[x]], x -> DirectedInfinity[1]]


      I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):



      $qquad frac s2 sqrtfrac ue$



      But for some reason, when using Limit, I get



      DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]


      So my questions are:



      What is going here?

      What issues should I be aware of when using Limit?










      share|improve this question









      New contributor




      Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I am trying to evaluate a limit:



      gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
      Limit[Re[gamma[x]], x -> DirectedInfinity[1]]


      I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):



      $qquad frac s2 sqrtfrac ue$



      But for some reason, when using Limit, I get



      DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]


      So my questions are:



      What is going here?

      What issues should I be aware of when using Limit?







      calculus-and-analysis complex






      share|improve this question









      New contributor




      Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 44 mins ago









      m_goldberg

      89.3k873200




      89.3k873200






      New contributor




      Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 4 hours ago









      VillaVilla

      1083




      1083




      New contributor




      Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:



          Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]



          (s u)/(2 Sqrt[e u])







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you, it worked.
            $endgroup$
            – Villa
            3 hours ago










          • $begingroup$
            Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
            $endgroup$
            – Bob Hanlon
            2 hours ago











          Your Answer








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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:



          Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]



          (s u)/(2 Sqrt[e u])







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you, it worked.
            $endgroup$
            – Villa
            3 hours ago










          • $begingroup$
            Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
            $endgroup$
            – Bob Hanlon
            2 hours ago















          4












          $begingroup$

          I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:



          Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]



          (s u)/(2 Sqrt[e u])







          share|improve this answer











          $endgroup$












          • $begingroup$
            Thank you, it worked.
            $endgroup$
            – Villa
            3 hours ago










          • $begingroup$
            Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
            $endgroup$
            – Bob Hanlon
            2 hours ago













          4












          4








          4





          $begingroup$

          I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:



          Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]



          (s u)/(2 Sqrt[e u])







          share|improve this answer











          $endgroup$



          I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:



          Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]



          (s u)/(2 Sqrt[e u])








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 mins ago









          m_goldberg

          89.3k873200




          89.3k873200










          answered 3 hours ago









          Carl WollCarl Woll

          76.3k3100200




          76.3k3100200











          • $begingroup$
            Thank you, it worked.
            $endgroup$
            – Villa
            3 hours ago










          • $begingroup$
            Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
            $endgroup$
            – Bob Hanlon
            2 hours ago
















          • $begingroup$
            Thank you, it worked.
            $endgroup$
            – Villa
            3 hours ago










          • $begingroup$
            Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
            $endgroup$
            – Bob Hanlon
            2 hours ago















          $begingroup$
          Thank you, it worked.
          $endgroup$
          – Villa
          3 hours ago




          $begingroup$
          Thank you, it worked.
          $endgroup$
          – Villa
          3 hours ago












          $begingroup$
          Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
          $endgroup$
          – Bob Hanlon
          2 hours ago




          $begingroup$
          Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
          $endgroup$
          – Bob Hanlon
          2 hours ago










          Villa is a new contributor. Be nice, and check out our Code of Conduct.









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          Villa is a new contributor. Be nice, and check out our Code of Conduct.












          Villa is a new contributor. Be nice, and check out our Code of Conduct.











          Villa is a new contributor. Be nice, and check out our Code of Conduct.














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