Weird result in complex limit












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):



$qquad frac s2 sqrt{frac ue}$



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.
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$endgroup$

















    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):



    $qquad frac s2 sqrt{frac ue}$



    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.







    $endgroup$















      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):



      $qquad frac s2 sqrt{frac ue}$



      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.







      $endgroup$




      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):



      $qquad frac s2 sqrt{frac ue}$



      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




      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 47 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

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          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












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          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 10 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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