Finding lim n → ∞ sin ⁡ ( π n + 2 π log ⁡ ( n ) n ) n log ⁡ ( n ) sin ⁡ ( π n )

Question

Finding       lim          n      →      ∞                  sin      ⁡              (                  π          n                +                              2            π            log            ⁡            (            n            )                                n                          )                    n      log      ⁡      (      n      )      sin      ⁡              (                  π          n                )

Braylon Perez · Accepted Answer

First,      lim    n              sin      ⁡              π        n                    1      n        =  π  ,so you might as well calculate      lim          n      →      ∞                          sin        ⁡        (                  π          n                +                              2            π            log            ⁡            (            n            )                                n                          )                    log        ⁡        n              .You have      |                            sin          ⁡          (                      π            n                    +                                    2              π              log              ⁡              (              n              )                                      n                                )                          log          ⁡          n                      |    ≤      1          log      ⁡      n            →                                               0  ,so the limit is zero.

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