Find the limit of sequence (complex numbers)











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How do I find the limit of
$$Z_n=nsin{frac{i}{n}}$$
I'd say it is 0, but the book says $i$.






limits complex-numbers







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    up vote
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    How do I find the limit of
    $$Z_n=nsin{frac{i}{n}}$$
    I'd say it is 0, but the book says $i$.






    limits complex-numbers







    share|cite|improve this question
























      up vote
      -3
      down vote

      favorite









      up vote
      -3
      down vote

      favorite











      How do I find the limit of
      $$Z_n=nsin{frac{i}{n}}$$
      I'd say it is 0, but the book says $i$.






      limits complex-numbers







      share|cite|improve this question













      How do I find the limit of
      $$Z_n=nsin{frac{i}{n}}$$
      I'd say it is 0, but the book says $i$.






      limits complex-numbers




      limits complex-numbers






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      share|cite|improve this question











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      asked 7 hours ago









      user3132457

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          $Z_n= ifrac {sin (i/n)} {i/n} to i$.






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            Hints:



            $$sin(ix) = icdot sinh b\lim_{xto0} frac{sinh(x)}{x} = 1$$






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              2 Answers
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              2 Answers
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              $Z_n= ifrac {sin (i/n)} {i/n} to i$.






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted










                $Z_n= ifrac {sin (i/n)} {i/n} to i$.






                share|cite|improve this answer























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                  accepted







                  up vote
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                  accepted






                  $Z_n= ifrac {sin (i/n)} {i/n} to i$.






                  share|cite|improve this answer












                  $Z_n= ifrac {sin (i/n)} {i/n} to i$.







                  share|cite|improve this answer












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                  answered 7 hours ago









                  Kavi Rama Murthy

                  39k31748




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                      up vote
                      0
                      down vote













                      Hints:



                      $$sin(ix) = icdot sinh b\lim_{xto0} frac{sinh(x)}{x} = 1$$






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        Hints:



                        $$sin(ix) = icdot sinh b\lim_{xto0} frac{sinh(x)}{x} = 1$$






                        share|cite|improve this answer























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                          up vote
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                          down vote









                          Hints:



                          $$sin(ix) = icdot sinh b\lim_{xto0} frac{sinh(x)}{x} = 1$$






                          share|cite|improve this answer












                          Hints:



                          $$sin(ix) = icdot sinh b\lim_{xto0} frac{sinh(x)}{x} = 1$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 7 hours ago









                          5xum

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