This morning a colleague asked me how I would solve quickly this limit, lim x → π 4 cos ⁡ ( 2 x ) 2 cos ⁡ ( x ) − 2

Question

This morning a colleague asked me how I would solve quickly this limit,      lim          x      →              π        4                        cos      ⁡      (      2      x      )              2      cos      ⁡      (      x      )      −              2

Sophia Mcdowell · Accepted Answer

Consider rewriting the numerator as   cos  ⁡  (  2  x  )  −  cos  ⁡  (      π    2    )  . Then we can use the same manipulation to get  cos  ⁡  (  2  x  )  −  cos  ⁡  (      π    2    )  =  −  2  sin  ⁡      (                  2        x        +                  π          2                    2        )    sin  ⁡      (                  2        x        −                  π          2                    2        )    =  −  2  sin  ⁡      (    x    +          π      4        )    sin  ⁡      (    x    −          π      4        )  Substituting this in gives us      lim          x      →              π        4                        −      2      sin      ⁡              (        x        +                  π          4                )            sin      ⁡              (        x        −                  π          4                )                    2              (        −        2        sin        ⁡                  (                                    x              +                              π                4                                      2                    )                sin        ⁡                  (                                    x              −                              π                4                                      2                    )                )              =      1    2        lim          x      →              π        4                        sin      ⁡              (        x        +                  π          4                )                    sin      ⁡              (                              x            +                          π              4                                2                )              ⋅      lim          x      →              π        4                        sin      ⁡              (        x        −                  π          4                )                    sin      ⁡              (                              x            −                          π              4                                2                )            supposing for the moment that those two limits exist. (also note the additional factor of 2 in the denominator which seems to have been dropped in your manipulation)For the first limit we can simply evaluate at   x  =      π    4   to get       1          1              2              =      2    .For the second limit, consider rewriting   sin  ⁡  (  x  −      π    4    ) in the numerator as   sin  ⁡      (    2                  x        −                  π          4                    2        )    =  2  sin  ⁡      (                  x        −                  π          4                    2        )    cos  ⁡      (                  x        −                  π          4                    2        )    . This gives us that      lim          x      →              π        4                        sin      ⁡              (        x        −                  π          4                )                    sin      ⁡              (                              x            −                          π              4                                2                )              =      lim          x      →              π        4              2  cos  ⁡      (                  x        −                  π          4                    2        )    =  2So our total limit is       1    2    ⋅      2    ⋅  2  =      2    , which agrees with your initial solution.

Wade Bullock · Accepted Answer

If you set   x  =  u  +      π    4   with   u  →  0 then you get  f  (  x  )  =                    cos        ⁡        (        2        x        )                    2        cos        ⁡        (        x        )        −                  2                      =                    −        2        sin        ⁡        (        u        )                                                            cos                ⁡                (                u                )                            ⏞                                            →            1                                                2                (                                                            cos                ⁡                (                u                )                −                1                            ⏟                                            ≪                         sin            ⁡            (            u            )                          −        sin        ⁡        (        u        )        )              ∼                    −        2        sin        ⁡        (        u        )                    −                  2                sin        ⁡        (        u        )              →      2  The fact that     cos  ⁡  (  u  )  −  1  ≪  sin  ⁡  (  u  ) can be inferred from:      |                            cos          ⁡          (          u          )          −          1                          sin          ⁡          (          u          )                      |    =      |                            2          sin          ⁡          (          u                      /                    2                      )            2                                    sin          ⁡          (          u          )                      |    =                              |                                                    sin                ⁡                (                u                                  /                                2                                  )                  2                                                                              u                  2                                                  /                                4                                              |                ⏟                    →      1        ×                              |                                    u                              sin                ⁡                (                u                )                                              |                ⏟                    →      1        ×                                          |                    8          u                      |                          ⏟                    →      0        →  0

This morning a colleague asked me how I would solve quickly this limit, <munder> <mo movable

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