Given lim x → 0 f ( 2 x ) − f ( x ) x = 0 and lim x → 0 f ( x ) = 0 show that lim x → 0 f ( x ) x = 0

Question

Given       lim          x      →      0                  f      (      2      x      )      −      f      (      x      )        x    =  0 and       lim          x      →      0        f  (  x  )  =  0 show that       lim          x      →      0                  f      (      x      )        x    =  0

Melina Richard · Accepted Answer

We could simplify all this. Let      ρ        (    x    )    =      sup                  |        y        |            ⩽              |        x        |                  |                  f                  (          2          y          )                −        f                  (          y          )                    y        |  We have for   n  ⩾  0                                          |            f                          (              x              )                        −            f                          (                              x                                                      2                                                        n                    +                    1                                                              )                        |                                      =                                          |                          ∑                              k                =                0                                            n                                                    (              f                              (                                  x                                                            2                                                              k                                                                      )                            −              f                              (                                  x                                                            2                                                              k                      +                      1                                                                      )                            )                        |                                                    ⩽                                          ∑                          k              =              0                                      n                                            ρ                                (            x            )                                              |              x              |                                                      2                                            k                +                1                                                                              ⩽                                          |            x            |                                ρ                                (            x            )                                (            1            −                          1                                                2                                                  n                  +                  1                                                      )                              By taking the limit of this inequality when   n  →  ∞, it follows that      |                  f                  (          x          )                    x        |    ⩽      ρ        (    x    )              ⟶              x      →      0        ⁡  0

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