Limit lim x → 0 1 ln ⁡ ( x + 1 ) − 1 x The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the O -notation.

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Limit       lim          x      →      0            1          ln      ⁡      (      x      +      1      )        −      1    x  The problem is I don&#039;t know if I can calculate it normally like with a change of variables or not. Keep in mind that I&#039;m not allowed to use L&#039;Hôpital&#039;s rule nor the       O  -notation.

Larry Yates · Accepted Answer

lim          x      →      0            1          ln      ⁡      (      x      +      1      )        −      1    x    =      lim          x      →      1            1          ln      ⁡      (      x      )        −      1          x      −      1        =  Lsince   x  →  1 is equivalent to       x    2    →  1, we can write  L  =      lim          x      →      1            1          ln      ⁡      (              x        2            )        −      1                  x        2            −      1        =      lim          x      →      1            1          2      ln      ⁡      (      x      )        +      1    2        (          1              x        +        1              −          1              x        −        1              )    =      1    2    L  +      lim          x      →      1            1    2        1          x      +      1        =      1    2    L  +      1    4  Hence,   L  =      1    2    L  +      1    4   and   L  =      1    2

Camille Flynn · Accepted Answer

I assume you know about the geometric series and Taylor&#039;s theorem to integrate them, but you might not. I apologize if you don&#039;t, as this will not be of much use.We have      1          log      ⁡      (      x      +      1      )        −      1    x    =            x      −      log      ⁡      (      x      +      1      )              x      log      ⁡      (      x      +      1      )        =            1      −                        log          ⁡          (          x          +          1          )                x                    log      ⁡      (      x      +      1      )        .Now      1          1      +      x        =      1          1      −      (      −      x      )        =      ∑          n      ≥      0        (  −  x      )    n    ,hence  log  ⁡  (  x  +  1  )  =      ∑          n      ≥      0                  −      (      −      x              )                  n          +          1                            n      +      1        =      ∑          n      ≥      0                  (      −      1              )        n                    x                  n          +          1                            n      +      1        =  x  −            x      2        2    +            x      3        3    −  ⋯  .Also,            log      ⁡      (      x      +      1      )        x    =      ∑          n      ≥      0                  (      −      1              )        n                    x        n                    n      +      1        ,hence  1  −            log      ⁡      (      x      +      1      )        x    =      ∑          n      ≥      1                  (      −      1              )                  n          +          1                            x        n                    n      +      1        =      x    2    −            x      2        3    +            x      3        4    −  ⋯  .Therefore,      lim          x      →      0            1          log      ⁡      (      x      +      1      )        −      1    x    =      lim          x      →      0                          x        2            −                        x          2                3            +                        x          3                4            −      ⋯              x      −                        x          2                2            +                        x          3                3            −      ⋯        =      lim          x      →      0                          1        2            −              x        3            +                        x          2                4            −      ⋯              1      −              x        2            +                        x          2                3            −      ⋯        =      1    2    .Note : you don&#039;t need the whole series expansion, you could approximate to the second term using Taylor&#039;s theorem. But I guess that falls into your &quot;big-O notation&quot; category of proofs.Hope that helps,

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