Evaluate lim x → 0 ( 1 ln ⁡ ( 1 + x ) + 1 ln ⁡ ( 1 − x ) )

Question

Evaluate             lim              x        →        0                    (              1                  ln          ⁡          (          1          +          x          )                    +              1                  ln          ⁡          (          1          −          x          )                    )

Josie123 · Accepted Answer

What&#039;s wrong with L&#039;Hopital&#039;s?                              lim                      x            →            0                                    1                      ln            ⁡            (            1            +            x            )                          +                  1                      ln            ⁡            (            1            −            x            )                                              =                  lim                      x            →            0                                                ln            ⁡            (            1            −                          x              2                        )                                ln            ⁡            (            1            +            x            )            ln            ⁡            (            1            −            x            )                                                            =                  lim                      x            →            0                                                −            2            x                          /                        (            1            −                          x              2                        )                                ln            ⁡            (            1            −            x            )                          /                        (            1            +            x            )            −            ln            ⁡            (            1            +            x            )                          /                        (            1            −            x            )                                                            =                  lim                      x            →            0                                                −            2            x                                (            1            −            x            )            ln            ⁡            (            1            −            x            )            −            (            1            +            x            )            ln            ⁡            (            1            +            x            )                                                            =                  lim                      x            →            0                                                −            2                                −            1            −            ln            ⁡            (            1            −            x            )            −            1            −            ln            ⁡            (            1            +            x            )                                                            =                              −            2                                −            2                                                            =        1.

skylsn · Accepted Answer

Since   log  ⁡  (  1  +  x  )  =  x  −            x      2        2    +  O  (      x    3    ), we have   log  ⁡  (  1  −  x  )  =  −  x  −            x      2        2    +  O  (      x    3    ). Therefore,                              1                      log            ⁡            (            1            +            x            )                          +                  1                      log            ⁡            (            1            −            x            )                                              =                  1                      x            (            1            −                          x              2                        +            O            (                          x              2                        )            )                          +                  1                      −            x            (            1            +                          x              2                        +            O            (                          x              2                        )            )                                                        (                      ∗                    )                                          =                              1            +                          x              2                        +            O            (                          x              2                        )                    x                −                              1            −                          x              2                        +            O            (                          x              2                        )                    x                                                  =                              x            +            O            (                          x              2                        )                    x                    where   (  ∗  ) is because       1          1      +      x        =  1  −  x  +  O  (      x    2    )Therefore,      lim          x      →      0            1          log      ⁡      (      1      +      x      )        +      1          log      ⁡      (      1      −      x      )        =  1

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