Let f : R → R be defined by f ( x ) = { x [ 1 + sin ⁡ ( ln ⁡ x ) ] if x &gt; 0 0 if x = 0 x + − x sin 2 ⁡ ( ln ⁡ | x | ) if x &lt; 0. Find the values of the four Dini derivatives of f at x = 0.I could easily find out the values of the upper right D + f and lower right D + f Dini derivatives at x = 0 , values being 2 and 0 respectively. But, in case of upper left Dini derivative of f at x = 0 , we have: D − f ( 0 ) = lim sup h → 0 − f ( h ) − f ( 0 ) h = 1 + lim sup h → 0 − − h sin 2 ⁡ ( ln ⁡ | h | ) h . Next putting: − h = p so that p → 0 + ,, I obtained: D − f ( 0 ) = 1 − lim inf p → 0 + | sin ⁡ ( ln ⁡ | p | ) | p .Since it is limit inferior and there is a modulus of sine function present inside, so can we take the sine part as 0 ? This way the answer would match but I feel like I'm avoiding the effect of the term p present in the denominator. I am confused. Please give some insights. Thanks in advance.

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Let   f  :      R    →      R   be defined by  f  (  x  )  =      {                            x          [          1          +          sin          ⁡          (          ln          ⁡          x          )          ]                          if           x          &amp;gt;          0                                      0                          if           x          =          0                                      x          +                      −            x                          sin              2                        ⁡            (            ln            ⁡                          |                        x                          |                        )                                    if           x          &amp;lt;          0.                        Find the values of the four Dini derivatives of   f at   x  =  0.I could easily find out the values of the upper right       D    +    f and lower right       D    +    f Dini derivatives at   x  =  0  , values being 2 and 0 respectively. But, in case of upper left Dini derivative of   f at   x  =  0  , we have:                              D          −                f        (        0        )                            =                  lim sup                      h            →            0            −                                                f            (            h            )            −            f            (            0            )                    h                                                                =        1        +                  lim sup                      h            →            0            −                                                −            h                          sin              2                        ⁡            (            ln            ⁡                          |                        h                          |                        )                    h                .            Next putting:   −  h  =  p so that   p  →  0  +  ,, I obtained:      D    −    f  (  0  )  =  1  −      lim inf          p      →      0      +                          |            sin      ⁡      (      ln      ⁡              |            p              |            )              |                    p        .Since it is limit inferior and there is a modulus of sine function present inside, so can we take the sine part as 0 ? This way the answer would match but I feel like I&#039;m avoiding the effect of the term       p   present in the denominator. I am confused. Please give some insights. Thanks in advance.

Savanah Hernandez · Accepted Answer

We have                              D                      −                          f        (        0        )                            =                  lim sup                      h            →                          0                              −                                                                          h            +                          −              h                              sin                2                            ⁡              (              ln              ⁡              (              −              h              )              )                                h                                                  =                  lim sup                      h            →                          0                              −                                                              (          1          −                      −                                                            sin                  2                                ⁡                (                ln                ⁡                (                −                h                )                )                            h                                )                    In the last line, we used       1    h    =  −            1              h        2             for   h  &amp;lt;  0. Let  g  (  h  )  =  1  −      −                            sin          2                ⁡        (        ln        ⁡        (        −        h        )        )            h      Clearly,   g  (  h  )  ≤  1 for all   h  &amp;lt;  0. The limit superior would then be equal to 1 if, for example, we show g attains the value of 1 in every left neighborhood of 0. Consider the sequence      x    n    =  −      e          −      n      π      Note       x    n    &amp;lt;  0,       lim          n      →      ∞            x    n    =  0, and  g  (      x    n    )  =  1  −            e              n        π                    sin      2        ⁡    (    −    n    π    )    =  1  −  0  =  1Thus,       D          −        f  (  0  )  =  1.

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