"use the squeeze theorem to evaluate the following limits: an=sin(1/n) over n. an=cos(1/n)-1 / 2^n"
an = sin(1/n) Try squeezing this between 1/n and 1/n2 (should get 0)
an = cos(1/n) - 1/2n Try squeezing this between 1 - 1/n and 1 - 1/2n (should get 1)
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