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 $i$.










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    up vote
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    favorite












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










    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 $i$.










      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






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      asked 7 hours ago









      user3132457

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      475






















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






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                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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                  down vote



                  accepted






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






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







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                  answered 7 hours ago









                  Kavi Rama Murthy

                  39k31748




                  39k31748






















                      up vote
                      0
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                      Hints:



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






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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























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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

                          88.2k392160




                          88.2k392160






























                               

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