How does one show that lim n → ∞ n 2 k ⋅ 1 − cos k ⁡ ( 2 π n ) = π?How does one show that L = π for k&gt;0?An attempt:For k=2 (2) lim n → ∞ n 2 ⋅ 1 − cos 2 ⁡ ( 2 π n ) = L (3) lim n → ∞ n 2 ⋅ sin ⁡ ( 2 π n ) = L (4) lim n → ∞ π ⋅ sin ⁡ ( 2 π n ) 2 π n = L (5) L = π

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How does one show that       lim          n      →      ∞                  n              2        k              ⋅      1    −          cos      k        ⁡          (                                    2            π                    n                    )        =  π?How does one show that   L  =  π for k&amp;gt;0?An attempt:For k=2                    (2)                              lim                      n            →            ∞                                                n            2                          ⋅                  1          −                      cos            2                    ⁡                      (                                                            2                  π                                n                                      )                          =        L                                (3)                              lim                      n            →            ∞                                                n            2                          ⋅        sin        ⁡                  (                                                    2                π                            n                                )                =        L                                (4)                              lim                      n            →            ∞                          π        ⋅                                            sin              ⁡                              (                                                                            2                      π                                        n                                                  )                                                                                      2                  π                                n                                                    =        L                                (5)                    L        =        π

vikafa4g · Accepted Answer

Hint. One may use a Taylor expansion of       cos    k    ⁡  x as   x  →  0, getting      cos    k    ⁡  x  =  1  −      k    2        x    2    +  O  (      x    4    )thus, as   x  →      0    +  ,      1    −          cos      k        ⁡    x    =            k      2        x  +  O  (      x    2    )then putting   x  =                    2        π            n      , as   n  →  ∞

hughy46u · Accepted Answer

x  →  0    cos  ⁡  x  =  1  −      1    2        x    2    +      1          4      !            x    4    +  .  .  .  ∼  1  −            1      2            x    2          cos    k    ⁡  x  ∼  (  1  −      1    2        x    2        )    k    ∼  1  −            k      2            x    2    now put into ,            2      π        n    →  0      lim          n      →      ∞                  n              2        k              ⋅      1    −          cos      k        ⁡          (                                    2            π                    n                    )        =        lim          n      →      ∞                  n              2        k              ⋅      1    −    (    1    −                  k        2              (                  2        π            n              )      2        )      =      lim          n      →      ∞                  n              2        k              ⋅                    k        2              (                  2        π            n              )      2          =      lim          n      →      ∞                  n              2        k              ⋅                    k        2              ×  (            2      π        n    )  =        lim          n      →      ∞                  1              2              ⋅                    1        2              ×  (            2      π        1    )  =                    2        π            2        =  π

How does one show that <munder> <mo movablelimits="true" form="prefix">lim <mrow class="M

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