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?
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?