How do I evaluate lim x → π / 2 sin ⁡ x − 1 ( 1 + cos ⁡ 2 x ) without using L'Hospital's rule?

Question

How do I evaluate       lim          x      →      π              /            2                  sin      ⁡      x      −      1              (      1      +      cos      ⁡      2      x      )       without using L&#039;Hospital&#039;s rule?

jugf5 · Accepted Answer

lim          x      →              π        2                        sin      ⁡      x      −      1              (      1      +      cos      ⁡      2      x      )        =      lim          x      →              π        2                        sin      ⁡      x      −      1              (      1      +              cos        2            ⁡      x      −              sin        2            ⁡      x      )        =    =      lim          x      →              π        2                        sin      ⁡      x      −      1              (      1      +      (      1      −              sin        2            ⁡      x      )      −              sin        2            ⁡      x      )        =    =      lim          x      →              π        2                        sin      ⁡      x      −      1              2      (      1      −              sin        2            ⁡      x      )        =    =      lim          x      →              π        2                        sin      ⁡      x      −      1              2      (      1      −      sin      ⁡      x      )      (      1      +      sin      ⁡      x      )        =    =      lim          x      →              π        2                        −      1              2      (      1      +      sin      ⁡      x      )        =  −      1    4    Using Taylor for       x    0    =      π    2  ...  sin  ⁡  (  x  )  =  1  −  (  π      /    2  −  x      )    2        /    2  +  …  cos  ⁡  (  2  x  )  =  −  1  +  2  (  π      /    2  −  x      )    2    +  …Then:      lim          x      →              π        2                        sin      ⁡      x      −      1              (      1      +      cos      ⁡      2      x      )        =      lim          x      →              π        2                        1      −      (      π              /            2      −      x              )        2                    /            2      −      1      +      …              (      1      −      1      +      2      (      π              /            2      −      x              )        2            +      …      )        =    =      lim          x      →              π        2                        −      (      π              /            2      −      x              )        2                    /            2      +      …              2      (      π              /            2      −      x              )        2            +      …        =  −      1    4

Patatiniuh · Accepted Answer

Use the relevant duplication formula:            sin      ⁡      x      −      1              1      +      cos      ⁡      2      x        =            sin      ⁡      x      −      1              2              cos        2            ⁡      x        =                    sin        2            ⁡      x      −      1              2              cos        2            ⁡      x      (      sin      ⁡      x      +      1      )        =            −              cos        2            ⁡      x              2              cos        2            ⁡      x      (      sin      ⁡      x      +      1      )        =  −      1          2      (      1      +      sin      ⁡      x      )        →  −      1    4    .

How do I evaluate <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX

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