The question is finding the value of L = lim x → 0 1 − cos ⁡ x cos ⁡ 2 x cos ⁡ 3 x x 2 if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .

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The question is finding the value of   L  =      lim          x      →      0                  1      −      cos      ⁡      x      cos      ⁡      2      x      cos      ⁡      3      x              x      2       if it exists . I&#039;ve found the answer using taylor&#039;s formula but I&#039;m looking for other solutions maybe using trigonometric identities .

victorollaa5 · Accepted Answer

Observe you have                    cos        ⁡        x        cos        ⁡        2        x        cos        ⁡        3        x        =        cos        ⁡        x        (        cos        ⁡        5        x        +        sin        ⁡        2        x        sin        ⁡        3        x        )            and                    cos        ⁡        x        cos        ⁡        5        x        =        cos        ⁡        6        x        +        sin        ⁡        x        sin        ⁡        5        x            which means                                          1            −            cos            ⁡            x            cos            ⁡            2            x            cos            ⁡            3            x                                x            2                          =                                                   1            −            cos            ⁡            6            x                                x            2                          −                              sin            ⁡            x            sin            ⁡            5            x                                x            2                          −                              cos            ⁡            x            sin            ⁡            2            x            sin            ⁡            3            x                                x            2                                              =                             18                                            sin              2                        ⁡            3            x                                (            3            x                          )              2                                      −        5                              sin            ⁡            x            sin            ⁡            5            x                                x            (            5            x            )                          −        6                              cos            ⁡            x            sin            ⁡            2            x            sin            ⁡            3            x                                (            2            x            )            (            3            x            )                              In the limit, we have18-5-6=7

vittorecostao1 · Accepted Answer

You can use the formulae  cos  ⁡  A  cos  ⁡  B  =      1    2    cos  ⁡  (  A  +  B  )  +      1    2    cos  ⁡  (  A  −  B  )  .then:                    cos        ⁡        (        x        )        cos        ⁡        (        2        x        )        cos        ⁡        (        3        x        )                            =        cos        ⁡        (        x        )        (        cos        ⁡        (        5        x        )        +        sin        ⁡        (        2        x        )        sin        ⁡        (        3        x        )        )                                          =        cos        ⁡        (        x        )        −        sin        ⁡        (        x        )        sin        ⁡        (        5        x        )        −        cos        ⁡        (        x        )        sin        ⁡        (        2        x        )        sin        ⁡        (        3        x        )        .            and using the remarkable limit       lim          x      →      0                  sin      ⁡      (      x      )        x    =  1.You get as first answer 18-5-6=7Or:You can directly apply L&#039;Hospital rule and get that:                                                          lim                      x            →            0                                                1            −            cos            ⁡            (            x            )            cos            ⁡            (            2            x            )            cos            ⁡            (            3            x            )                                x            2                                                            =                  lim                      x            →            0                                    1          2                                      cos            ⁡            (            2            x            )            cos            ⁡            (            3            x            )            sin            ⁡            (            x            )            +            2            cos            ⁡            (            x            )            cos            ⁡            (            3            x            )            sin            ⁡            (            2            x            )            +            3            cos            ⁡            (            x            )            cos            ⁡            (            2            x            )            sin            ⁡            (            3            x            )                    x                                                  =                              1            +            4            +            9                    2                =                  14          2                =        7.

The question is finding the value of L = <munder> <mo movablelimits="true" form="pref

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