The answer of the following limit: lim x → 0 ( 1 ln ⁡ cos ⁡ ( x ) + 2 sin 2 ⁡ ( x ) )

Question

The answer of the following limit:      lim          x      →      0            (          1              ln        ⁡        cos        ⁡        (        x        )              +          2                        sin                      2                          ⁡        (        x        )              )

Daniel Valdez · Accepted Answer

It is because when you expand   cos  ⁡  (  x  ) and   ln  ⁡  (  cos  ⁡  (  x  )  ), you need to consider more fourth-order term. Specifically,  cos  ⁡  (  x  )  =  1  −            x      2        2    +            x      4        24    +  o  (      x    5    )  .Then  ln  ⁡      (    cos    ⁡    (    x    )    )    =  −            x      2        2    +            x      4        24    −                    (        −                              x            2                    2                +                              x            4                    24                )            2        2    +  o  (      x    5    )  =  −            x      2        2    −            x      4        12    +  o  (      x    5    )  .With the same expression for   sin  ⁡  (  x  ) as you have written, we obtain                              1                      ln            ⁡                          (              cos              ⁡              (              x              )              )                                      +                  2                                    sin              2                        ⁡            (            x            )                                              =                  1                      −                                          x                2                            2                        −                                          x                4                            12                        +            o            (                          x              5                        )                          +                  2                                    x              2                        −                                          x                4                            3                        +            o            (                          x              4                        )                                                            =                                            x              2                        −                                          x                4                            3                        −                          x              2                        −                                          x                4                            6                                            −                                          x                4                            2                        +            o            (                          x              4                        )                                                            =                              x            4                                              x              4                        +            o            (                          x              4                        )                          .

Kyla Ayers · Accepted Answer

We need more terms to obtain  ln  ⁡  (  cos  ⁡  x  )  =  ln  ⁡      (    1    −                  x                  2                    2        +                  x                  4                    24        +    o          (              x                  4                    )        )    =  −            x              2              2    −            x              4              12    +  o      (          x              4              )  and then by binomial expansion      1          −                        x                      2                          2            −                        x                      4                          12            +      o              (                  x                      4                          )              +      2                  x                  2                    −                        x                      4                          3            +      o              (                  x                      4                          )              =      2          x      2            (    −          1              1        +                              x            2                    6                +        o                  (                      x                          2                                )                      +          1              1        −                              x                          2                                3                +        o                  (                      x                          2                                )                      )    =      2          x      2            (    −    1    +                  x        2            6        +    1    +                  x                  2                    3        +    o          (              x                  2                    )        )    =      2          x      2            (                  x        2            2        +    o    (          x      2        )    )    =  1  +  o  (  1  )  →  1

The answer of the following limit: <munder> <mo movablelimits="true" form="prefix">lim <

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