I need help finding this limit: lim x → 1 sin ⁡ π x sin ⁡ 3 π x

Question

I need help finding this limit:      lim          x      →      1                  sin      ⁡              π        x                    sin      ⁡              3        π        x

Caiden Barrett · Accepted Answer

Recall that  sin  ⁡  (  A  +  B  )  =  sin  ⁡  A  cos  ⁡  B  +  cos  ⁡  A  sin  ⁡  BWe have                              lim                      x            →            1                                                sin            ⁡                          π              x                                            sin            ⁡                          3              π              x                                                          =                  lim                      x            →            1                                                sin            ⁡                          π              x                                            sin            ⁡            (            2            π            x            +            π            x            )                                                            =                  lim                      x            →            1                                                sin            ⁡                          π              x                                                          sin              ⁡              (              2              π              x              )                        cos            ⁡            (            π            x            )            +            cos            ⁡            (            2            π            x            )            sin            ⁡            (            π            x            )                                                            =                  lim                      x            →            1                                                sin            ⁡                          π              x                                                          2              sin              ⁡              (              π              x              )              cos              ⁡              (              π              x              )                        cos            ⁡            (            π            x            )            +            cos            ⁡            (            2            π            x            )            sin            ⁡            (            π            x            )                                                            =                  lim                      x            →            1                                    1                      2            cos            ⁡            (            π            x            )            cos            ⁡            (            π            x            )            +            cos            ⁡            (            2            π            x            )                                                            =                  1                      2            cos            ⁡            (            π            )            cos            ⁡            (            π            )            +            cos            ⁡            (            2            π            )                                                            =                  1          3

aggierabz2006zw · Accepted Answer

lim          x      →              1                        sin      ⁡              π                  x                            sin      ⁡              3        π                  x                      .To solve this problem I recommend to use the Taylor series:                    (1)                    sin        ⁡        (        x        )        =                  (          x          −                                    x              3                                      3              !                                +                                    x              5                                      5              !                                +                                    x              7                                      7              !                                +          .          .          .          .          )                =                  ∑                      n            =            0                    ∞                                      (            −            1                          )              n                                      x                              2                n                +                1                                                          (            2            n            +            1            )            !                          .                                (2)                    sin        ⁡        (                  α                +                  π                )        =        −        sin        ⁡        (                  α                )        .            we need to perform a substitution   x  =  (  y  +  1  ), therefore,   y  →      0  Using the very well known formula (2) and the substitution   x  =  (  y  +  1  ) relatively our problem, therefore, we will have:      lim          y      →              0                        sin      ⁡              π                  y                            sin      ⁡              3        π                  y                      =      lim          y      →              0                        (      π              y            −                        (          π                      y                                )            3                                    3          !                    +                        (          π                      y                                )            5                                    5          !                    +                        (          π                      y                                )            7                                    7          !                    +      .      .      .      .      )              (      3      π              y            −                        (          3          π                      y                                )            3                                    3          !                    +                        (          3          π                      y                                )            5                                    5          !                    +                        (          3          π                      y                                )            7                                    7          !                    +      .      .      .      .      )        =    =      lim          y      →              0                        y              (        π        −                                            π                              3                                                    y                              2                                                          3            !                          +                                            π                              5                                                    y                              4                                                          5            !                          +                                            π                              7                                                    y                              6                                                          7            !                          +        .        .        .        .        )                    y              (        3        π        −                              (            3            π                          )                              3                                                    y                              2                                                          3            !                          +                              (            3            π                          )                              5                                                    y                              4                                                          5            !                          +                              (            3            π                          )                              7                                                    y                              6                                                          7            !                          +        .        .        .        .        )              =      lim          y      →              0                  π          3      π        =      1    3    .Therefore,      lim          x      →              1                        sin      ⁡              π                  x                            sin      ⁡              3        π                  x                      =      1    3    .

I need help finding this limit: <munder> <mo movablelimits="true" form="prefix">lim <mro

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New Questions in Calculus 1