Question
limn→∞n.cos(π4n).sin(π4n) is equal to
Answer: Option A
:
A
limx→∞n.cos(π4n)sin(π4n)
=limn→∞n2sinπ2n
=π4
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:
A
limx→∞n.cos(π4n)sin(π4n)
=limn→∞n2sinπ2n
=π4
Was this answer helpful ?
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