Sin (n) oscillates between -1 and 1
Therefore, Sin (n) / n oscillates between -1/n and 1/n
as n goes to infinity Sin (n) is between -1 or 1 while 1/n goes to 0
Therefore, lim n-> infinity sin(n) / n = 0
Ana A.
asked 03/26/22Find the limit of the sequence an={sin(n)/n}, if it exists
Sin (n) oscillates between -1 and 1
Therefore, Sin (n) / n oscillates between -1/n and 1/n
as n goes to infinity Sin (n) is between -1 or 1 while 1/n goes to 0
Therefore, lim n-> infinity sin(n) / n = 0
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