to show the limit as n-> infinity of sin(n)/n= 0 absolute value( sin(n)/n - 0) = abs( sin(n)/n) < or equal to 1/n < or equal to 1/n0 < epsilon so we can choose n0 by the Archimedeian principle is the way I handle the absolute value correct?
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Originally Posted by mtlchris to show the limit as n-> infinity of sin(n)/n= 0 absolute value( sin(n)/n - 0) = abs( sin(n)/n) < or equal to 1/n < or equal to 1/n0 < epsilon so we can choose n0 by the Archimedeian principle is the way I handle the absolute value correct? Looks fine to me...IF you can prove, or if you were given, that ... Tonio
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