Calculating lim x → π 6 ( sin ⁡ ( x − π 6 ) 3 2 − cos ⁡ x ) , without using L'Hospital rule

Question

Calculating       lim          x      →              π        6                  (                  sin        ⁡        (        x        −                  π          6                )                                                        3                                2                −        cos        ⁡        x              )  , without using L&#039;Hospital rule

Perman7z · Accepted Answer

Let put   t  =  x  −      π    6  we will compute      lim          t      →      0                  sin      ⁡      (      t      )                                3                2            −      cos      ⁡      (      t      +              π        6            )        =      lim          t      →      0                  sin      ⁡      (      t      )                                3                2            (      1      −      cos      ⁡      (      t      )      )      +                        sin          ⁡          (          t          )                2              =      lim          t      →      0            1                  1        2            +                        3                2                                          t            2                    (          1          −          cos          ⁡          (          t          )          )                                      t            2                    sin          ⁡          (          t          )                      =  2.

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