How to evaluate the limit lim x → 0 1 − cos ⁡ ( 4 x ) sin 2 ⁡ ( 7 x )

Question

How to evaluate the limit       lim          x      →      0                  1      −      cos      ⁡      (      4      x      )                      sin        2            ⁡      (      7      x      )

Kroatujon3 · Accepted Answer

Notice,      lim          x      →      0                  1      −      cos      ⁡      (      4      x      )                      sin        2            ⁡      (      7      x      )            lim          x      →      0                  1      −              cos        2            ⁡      (      4      x      )                      sin        2            ⁡      (      7      x      )      (      1      +      cos      ⁡      (      4      x      )      )        =      lim          x      →      0                          sin        2            ⁡      (      4      x      )                      sin        2            ⁡      (      7      x      )      (      1      +      cos      ⁡      (      4      x      )      )        =      lim          x      →      0                          sin        2            ⁡      (      4      x      )                      sin        2            ⁡      (      7      x      )        ⋅      lim          x      →      0            1          1      +      cos      ⁡      (      4      x      )        =      lim          x      →      0                  (                        sin          ⁡          (          4          x          )                          sin          ⁡          (          7          x          )                    )        2    ⋅      1          1      +      1        =      1    2        lim          x      →      0                  (              4        7                                          sin            ⁡            (            4            x            )                                4            x                                                sin            ⁡            (            7            x            )                                7            x                              )        2    =      1    2        16    49        lim          x      →      0                  (                                    sin            ⁡            (            4            x            )                                (            4            x            )                                                sin            ⁡            (            7            x            )                                (            7            x            )                              )        2    =      8    49              (              1        1            )        2    =            8      49

gaiaecologicaq2 · Accepted Answer

Hint:      lim          t      →      0                          sin        2            ⁡      (      4      t      )                      sin        2            ⁡      (      7      t      )        =      lim          t      →      0                          (                              sin            ⁡            4            t                    t                )            2                      (                              sin            ⁡            7            t                    t                )            2        =            4      2              7      2        ×                    (                  lim                      t            →            0                                                sin            ⁡            4            t                                4            t                          )            2                      (                  lim                      t            →            0                                                sin            ⁡            7            t                                7            t                          )            2        =      16    49    ×            1      2              1      2      Also                              lim                      t            →            0                                    1                      1            +            cos            ⁡            (            4            t            )                                              =                  1                      1            +            1                          =                  1          2

How to evaluate the limit <munder> <mo movablelimits="true" form="prefix">lim <mrow cl

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