I was looking up to some previous papers of a competitive exam when I found the following problem: lim n → ∞ ( 2 n + n ⋅ 2 n sin 2 ⁡ n 2 ) ( 1 2 n − n cos ⁡ 1 n )

Question

I was looking up to some previous papers of a competitive exam when I found the following problem:      lim          n      →      ∞                  (              2        n            +      n      ⋅              2        n                    sin        2            ⁡                        n          2                    )                      (                  1                      2            n            −            n            cos            ⁡                                          1                n                                                    )

grubijanebb · Accepted Answer

Your method is incorrect since      lim          n      →      ∞        1  +  n  ⋅      sin    2    ⁡            n      2        ≠  1So you cannot possibly use the       1          ∞       formTo solve the limit we can use logarithms  L  =      lim          n      →      ∞                  (      1      +      n      ⋅              sin        2            ⁡                        n          2                    )                      (                  1                      2            n            −            n            cos            ⁡                                          1                n                                                    )              ln  ⁡  L  =      lim          n      →      ∞                  ln      ⁡              (        1        +        n        ⋅                  sin          2                ⁡                              n            2                          )                    2      n      −      n      cos      ⁡                        1          n                      =      lim          n      →      ∞                  ln      ⁡              (        1        +        n        ⋅                  sin          2                ⁡                              n            2                          )              n  Since       sin    2    ⁡  x  ∈  [  0  ,  1  ]      lim          n      →      ∞                  ln      ⁡              (        1        )              n    ≤      lim          n      →      ∞                  ln      ⁡              (        1        +        n        ⋅                  sin          2                ⁡                              n            2                          )              n    ≤      lim          n      →      ∞                  ln      ⁡              (        1        +        n        )              n    0  ≤  ln  ⁡  L  ≤  0Therefore the limit is indeed 2

I was looking up to some previous papers of a competitive exam when I found the following problem:

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New Questions in Calculus 1