Problem in the properties of limit: lim x → π 3 sin ⁡ ( x − π 3 ) 1 − 2 cos ⁡ ( x ) I used the following property: if lim x → π 3 f ( x ) = Lthen lim x → π 3 1 f ( x ) = 1 L where L is a real number and nonzero,hence we have: lim x → π 3 1 − 2 cos ⁡ ( x ) sin ⁡ ( x − π 3 ) substititute x − π 3 = u: lim u → 0 1 − 2 cos ⁡ ( u + π 3 ) sin ⁡ ( u ) = lim u → 0 1 − cos ⁡ ( u ) + 2 sin ⁡ ( u ) sin ⁡ ( u ) = lim u → 0 1 − cos ⁡ ( u ) sin ⁡ ( u ) + 2 = lim u → 0 sin ⁡ ( u ) 1 + cos ⁡ ( u ) + 2 = 2 hence the main limit should be 1 2 which is wrong, but I don't know why, also is there any way to solve the problem without using Taylor series or L'hopital's rule?

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Problem in the properties of limit:       lim          x      →              π        3                        sin      ⁡              (        x        −                  π          3                )                    1      −      2      cos      ⁡              (        x        )            I used the following property: if      lim                  x        →                  π          3                      f  (  x  )  =  Lthen      lim          x      →              π        3                  1          f              (        x        )              =      1    L  where L is a real number and nonzero,hence we have:      lim                  x        →                  π          3                                1      −      2      cos      ⁡              (        x        )                    sin      ⁡              (        x        −                  π          3                )            substititute   x  −      π    3    =  u:      lim                  u        →        0                        1      −      2      cos      ⁡              (        u        +                  π          3                )                    sin      ⁡              (        u        )              =      lim                  u        →        0                        1      −      cos      ⁡              (        u        )            +              2            sin      ⁡              (        u        )                    sin      ⁡              (        u        )              =      lim                  u        →        0                        1      −      cos      ⁡              (        u        )                    sin      ⁡              (        u        )              +      2    =      lim                  u        →        0                        sin      ⁡              (        u        )                    1      +      cos      ⁡              (        u        )              +      2    =      2  hence the main limit should be       1          2       which is wrong, but I don&#039;t know why, also is there any way to solve the problem without using Taylor series or L&#039;hopital&#039;s rule?

Kristina Petty · Accepted Answer

Your derivation is absolutely fine and right, but we have that  1  −  2  cos  ⁡      (    u    +          π      3        )    =  1  −  2      1    2    cos  ⁡  u  +  2            3        2    sin  ⁡  u  =  1  −  cos  ⁡  u  +            3        sin  ⁡  utherefore      lim          x      →              π        3                        sin      ⁡              (        x        −                  π          3                )                    1      −      2      cos      ⁡              (        x        )              =      1          3      Note also that we don&#039;t need to invert the expression, indeed in the same way we have      lim                  u        →        0                        sin      ⁡              (        u        )                    1      −      2      cos      ⁡              (        u        +                  π          3                )              =      lim                  u        →        0                        sin      ⁡              (        u        )                    1      −      cos      ⁡      u      +              3            sin      ⁡      u        =      lim                  u        →        0                  1                            1          −          cos          ⁡          u                          sin          ⁡          u                    +              3              =      1          3      since            1      −      cos      ⁡      u              sin      ⁡      u        =  u            1      −      cos      ⁡      u              u      2            u          sin      ⁡      u        →  0

Problem in the properties of limit: <munder> <mo form="prefix">lim <mrow class="MJX-TeXAt

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