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use the squeeze theorem to evaluate the following limits: an=sin(1/n) / n. an=cos(1/n)-1 / 2^n

medlockb1234  Oct 8, 2017
edited by medlockb1234  Oct 9, 2017
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"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)

.

Alan  Oct 8, 2017

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