I used all the trig and log tricks but still can't compute this limit lim x → 0 ln ⁡ ( 1 + sin 2 ⁡ ( 2 x ) ) 1 − cos 2 ⁡ ( x ) .

Question

I used all the trig and log tricks but still can&#039;t compute this limit      lim          x      →      0                  ln      ⁡              (        1        +                  sin          2                ⁡        (        2        x        )        )                    1      −              cos        2            ⁡      (      x      )        .

Bornejecbo · Accepted Answer

Just using the standard limit       lim          t      →      0                  ln      ⁡      (      1      +      t      )        t    =  1 you get                                          ln            ⁡                          (              1              +                              sin                2                            ⁡              (              2              x              )              )                                            1            −                          cos              2                        ⁡            (            x            )                                              =                                          ln            ⁡                          (              1              +                              sin                2                            ⁡              (              2              x              )              )                                                          sin              2                        ⁡            (            2            x            )                          ⋅                                            sin              2                        ⁡            (            2            x            )                                              sin              2                        ⁡            x                                                            =                    4                  cos          2                ⁡        x        ⋅                              ln            ⁡                          (              1              +                              sin                2                            ⁡              (              2              x              )              )                                                          sin              2                        ⁡            (            2            x            )                                                                                                ⟶                                      x              →              0                                                  4        ⋅        1        ⋅        1        =        4

gnatopoditw · Accepted Answer

We know that (by asymptotic relations):  ln  ⁡  (  1  +  f  (  x  )  )  ∼  f  (  x  )        x  →  α      f  (  x  )  →  0Also:  sin  ⁡  (  x  )  ∼  t      x  →  0And:  cos  ⁡  (  x  )  ∼  1  −      1    2        x    2        x  →  0So, your limit is going to be:      lim          x      →      0                  ln      ⁡              (        1        +                  sin          2                ⁡        (        2        x        )        )                    1      −              cos        2            ⁡      (      x      )            ∼          lim          x      →      0                          sin        2            ⁡      (      2      x      )              1      −              cos        2            ⁡      (      x      )                lim          x      →      0                  4              x        2                    1      −              cos        2            ⁡      (      x      )        =      lim          x      →      0                  4              x        2                    (      1      +      cos      ⁡      (      x      )      )      ⋅      (      1      −      cos      ⁡      (      x      )      )                lim          x      →      0                  4              x        2                    2      ⋅              1        2                    x        2              =  4

I used all the trig and log tricks but still can't compute this limit <munder> <mo movableli

Answered question

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New Questions in Calculus 1