How can we compute the following limit not using derivative? lim x → 0 1 − cos ⁡ ( 3 x ) 5 x 2 I know lim x → 0 cos ⁡ ( x ) − 1 x = 0. But lim x → 0 1 − cos ⁡ ( 3 x ) 5 x 2 = lim x → 0 cos ⁡ ( 3 x ) − 1 3 x ( − 3 ) 5 x 2 = 0. − ∞

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How can we compute the following limit not using derivative?      lim          x      →      0                  1      −      cos      ⁡      (      3      x      )              5              x        2            I know       lim          x      →      0                  cos      ⁡      (      x      )      −      1        x    =  0. But       lim          x      →      0                  1      −      cos      ⁡      (      3      x      )              5              x        2              =      lim          x      →      0                  cos      ⁡      (      3      x      )      −      1              3      x                  (      −      3      )              5              x        2              =  0.  −  ∞

Bruce Bridges · Accepted Answer

lim          x      →      0                          1        −        cos        ⁡        3        x                    5                  x          2                      ⋅                    1        +        cos        ⁡        3        x                    1        +        cos        ⁡        3        x              =      lim          x      →      0                                    sin          2                ⁡        3        x                    5                  x          2                      ⋅      lim          x      →      0                  1              1        +        cos        ⁡        3        x              =      lim          x      →      0                                    sin          2                ⁡        3        x                    (        3        x                  )          2                      ⋅      lim          x      →      0                  9              5        +        5        cos        ⁡        3        x              =  1  ⋅            9              5        +        5              =            9      10

Isaiah Farrell · Accepted Answer

Hint:      sin    2    ⁡      (          3      2        x    )    =            1      −      cos      ⁡      3      x        2  and      lim          x      →      0                                    sin          2                ⁡        a        x                    (        a        x                  )          2                      =  1

How can we compute the following limit not using derivative? <munder> <mo movablelimits="

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