If cos−1⁡1n<π2, then limn→∞ (n+1)2πcos−1⁡1n−n is equal to

If cos11n<π2, then limn(n+1)2πcos11nn is equal to

  1. A

    2-ππ

  2. B

    π-2π

  3. C

    1

  4. D

    0

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    Solution:

       limn(n+1)2πcos11nn

    =limn2π(n+1)cos11nπ2n=limn2πncos11nπ2+cos11n=limn2πnsin11n+cos11n

    =limn2πsin11n1n+cos11n=2π1+π2=π2π

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