I'm looking for a solution to the following limit from above: lim x → 1 4 x 3 − 2 x x 2 − 1 ln ⁡ ( x + x 2 − 1 ) − 2 x 2 x 4 − 1

Question

I&#039;m looking for a solution to the following limit from above:      lim          x      →      1                                    4                      x            3                    −          2          x                                      x            2                    −          1                    ln      ⁡      (      x      +                        x          2                −        1            )      −      2              x        2                            x        4            −      1

Elliana Shelton · Accepted Answer

To make things a little bit easier, first factor x (whose limit is 1) out of the numerator, then use       x    4    −  1  =  (  x  −  1  )  (  x  +  1  )  (      x    2    +  1  ) for the denominator:                              lim                      x            →            1                                                                              4                                  x                  3                                −                2                x                                                              x                  2                                −                1                                      ln            ⁡            (            x            +                                          x                2                            −              1                        )            −            2                          x              2                                                          x              4                        −            1                                              =                  lim                      x            →            1                                                                              4                                  x                  2                                −                2                                                              x                  2                                −                1                                      ln            ⁡            (            x            +                                          x                2                            −              1                        )            −            2            x                                              x              4                        −            1                                                            =                  1          4                          lim                      x            →            1                                                                              4                                  x                  2                                −                2                                                              x                  2                                −                1                                      ln            ⁡            (            x            +                                          x                2                            −              1                        )            −            2            x                                x            −            1                              Denote   t  =  x  +            x      2        −    1  Using   ln  ⁡  (  1  +  x  )  =  x  −            x      2        2    +            x      3        3    +  o  (      x    3    ) we can write  ln  ⁡  (  x  +            x      2        −    1    )  =  ln  ⁡  (  1  +  (  t  −  1  )  )  =  (  t  −  1  )  −            (      t      −      1              )        2              2    +            (      t      −      1              )        3              3    +  o      (    (    t    −    1          )      3        )    .Notice that   o      (    (    t    −    1          )      3        )    =  o      (    (    x    −    1          )              3                  /                2              )   so that substituting this back into the limit we get      lim          x      →      1                                    4                      x            2                    −          2                                      x            2                    −          1                            (        x        +                              x            2                    −          1                −        1        −                              (            x            +                                          x                2                            −              1                        −            1                          )              2                                2                +                              (            x            +                                          x                2                            −              1                        −            1                          )              3                                3                +        o                  (          (          x          −          1                      )                          3                              /                            2                                )                )            −      2      x              x      −      1      where I put the factor       1    4   aside for now. We have      lim          x      →      1                                    4                      x            2                    −          2                                      x            2                    −          1                    o              (        (        x        −        1                  )                      3                          /                        2                          )                    x      −      1        =  0and this is why we had to take 3 terms from the Taylor series. So now we can consider just      lim          x      →      1                                    4                      x            2                    −          2                                      x            2                    −          1                            (        x        +                              x            2                    −          1                −        1        −                              (            x            +                                          x                2                            −              1                        −            1                          )              2                                2                +                              (            x            +                                          x                2                            −              1                        −            1                          )              3                                3                )            −      2      x              x      −      1      which can be solved using standard techniques: simplifying we get      2    3        lim          x      →      1                  8              x        4            −      10              x        3            −      2              x        2            +      5      x      −      1      +      2                        x          2                −        1            (      4              x        3            −      5              x        2            +      x      +      4      )                      x        2            −      1      which is the same as            2        3        lim          x      →      1                  8              x        4            −      10              x        3            −      2              x        2            +      5      x      −      1      +      2                        x          2                −        1            (      4              x        3            −      5              x        2            +      x      +      4      )              x      −      1      putting             2        3   aside and using L&#039;Hopital&#039;s rule we get      lim          x      →      1        2      x    −    1        (    32          x      3        −    30          x      2        −    4    x    +    5    +    2                  x        2            −      1        (    12          x      2        −    10    x    +    1    )        +                  2        x        (        4                  x          3                −        5                  x          2                +        x        +        4        )                              x          2                −        1              )  Bringing back the constants we took out, the original limit equals       1    4    ⋅            2        3    ⋅      16          2        =      4    3

I'm looking for a solution to the following limit from above: <munder> <mo movablelimits="tr

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