Find the limit of: lim x → π 3 1 − 2 cos ⁡ x π − 3 x

Question

Find the limit of:      lim          x      →              π        3                        1      −      2      cos      ⁡              x                    π      −      3      x

AlokMoopisppf · Accepted Answer

Use l&#039;Hôpital&#039;s rule:      lim          x      →              π        3                        1      −      2      cos      ⁡      (      x      )              π      −      3      x        =      lim          x      →              π        3                        2      sin      ⁡      (      x      )              −      3        =  −      2    3        lim          x      →              π        3              sin  ⁡  (  x  )  =  −      2    3    ⋅  sin  ⁡      (          π      3        )    =  −      2    3    ⋅            3        2    =  −      1          3

tomekmusicd9 · Accepted Answer

Put   x  =      π    3    +  h. Then the given limit is                              lim                      h            →            0                                                1            −            2            cos            ⁡                          (                              π                3                            +              h              )                                            −            3            h                                              =                  lim                      h            →            0                                                1            −            cos            ⁡            h            +                          3                        sin            ⁡            h                                −            3            h                                                            =                  lim                      h            →            0                                                2                          sin              2                        ⁡                          h              2                                            −            3            h                          +                  lim                      h            →            0                                    3                                      sin            ⁡            h                                −            3            h                                                            =        −                  1                      3

Find the limit of: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="

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