Can someone please solve this limit? lim x → 0 1 − cos ⁡ 2 π x x 2

Question

Can someone please solve this limit?      lim          x      →      0                  1      −      cos      ⁡      2      π      x              x      2

Arturo Wallace · Accepted Answer

We will use here L&#039;Hopital&#039;s rule, because our limit has a look       [          0      0        ]  , therefore,      lim          x      →              0                        (      1      −      cos      ⁡      2      π              x                    )                                          ′                                      (              x                  2                            )                                          ′                                =      lim          x      →              0                        2      π      sin      ⁡      (      2      π              x            )              2      x        .After the first differentiation our limit has a look       [          0      0        ]   again, therefore, we must get the second derivative, using L&#039;Hopital&#039;s rule, mentioned above:      lim          x      →              0                        (      π      sin      ⁡      (      2      π              x            )              )                                          ′                                      x                                  ′                      =      lim          x      →              0              (  2            π              2        cos  ⁡  (  2  π      x    )  )  =  2            π              2        .After the twice using differentiation we will receive:            lim              x        →        0                            1        −        cos        ⁡        2        π        x                    x        2              =    2                  π                    2              .

Daphne Haney · Accepted Answer

lim          x      →      0                  1      −      cos      ⁡      2      π      x              x      2        ,    cos  ⁡  (  2  x  )  =      cos          2        ⁡  (  x  )  −      sin          2        ⁡  (  x  )  ,        lim          x      →      0                  1      −      cos      ⁡      2      π      x              x      2        =      lim          x      →      0                  1      −              cos                  2                    ⁡      π              x            +              sin                  2                    ⁡      π              x                    x      2        =      lim          x      →      0                  2              sin                  2                    ⁡      π              x                    x      2        .We know the famous formula             lim              x        →                  0                                    sin        ⁡        (        x        )            x        =    1  , therefore,      lim          x      →      0                  2              sin                  2                    ⁡      π              x                    x      2        ⋅            π              2                    π              2              =      lim          x      →      0            (    2          π              2              ⋅                  sin        ⁡        π                  x                    x        ⋅                  sin        ⁡        π                  x                    x        )    =  2      π          2        .            lim              x        →        0                            1        −        cos        ⁡        2        π        x                    x        2              =    2                  π                    2              .

Can someone please solve this limit? <munder> <mo movablelimits="true" form="prefix">lim

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