What is the limit of the following function, as x approaches 1? lim x → 1 ln ⁡ ( x ) arctan ⁡ ( 1 − x ) .

Question

What is the limit of the following function, as x approaches 1?      lim          x      →              1                        ln      ⁡      (      x      )              arctan      ⁡      (      1      −      x      )        .

Lilliana Burton · Accepted Answer

Solution:      lim          x      →      1            (                  l        n                  (          x          )                            arctan        ⁡                  (          1          −          x          )                      )    =      lim          x      →      1            (                            d                      d            x                                    (          l          n                      (            x            )                    )                                      d                      d            x                                    (          arctan                                (            1            −            x            )                    )                      )    =      lim          x      →      1            (                  (                  1          x                )                    (                              −            1                                                              (                1                −                x                )                            2                        +            1                          )              )    =      lim          x      →      1            (                  −                  x          2                +        2        x        −        2            x        )    =      (                  −        (        1                  )          2                +        2        (        1        )        −        2                    (        1        )              )    =            −      1        1    =  −  1

Kapalci · Accepted Answer

lim          x      →              1                        ln      ⁡      (      x      )              arctan      ⁡      (      1      −      x      )        .We must make the substitution here   x  =  (  y  +  1  ), therefore      lim          y      →              0                        ln      ⁡      (      y      +      1      )              arctan      ⁡      (      −      y      )        =      lim          y      →              0                        ln      ⁡      (      y      +      1      )              −      arctan      ⁡      (      y      )        .I recommend to use here the Taylor series:                    (1)                    ln        ⁡        (        1        +        x        )        =        x        −                              x            2                    2                +                              x            3                    3                −        .        .        .        .                                (2)                    arctan        ⁡        (        x        )        =        x        −                              x            3                    3                +                              x            5                    5                −        .        .        .        .        ,                for                   −          1          ≤                      x                    ≤                      1                          ,        x        ≠        ±                  i                .            Therefore,      lim          y      →              0                        ln      ⁡      (      y      +      1      )              −      arctan      ⁡      (      y      )        =      lim          y      →              0                        (      y      −                        y          2                2            +                        y          3                3            −      .      .      .      .      )              −              (        y        −                              y            3                    3                +                              y            5                    5                −        .        .        .        .        )              =    =      lim          y      →              0                        y              (        1        −                  y          2                +                              y            2                    3                −        .        .        .        .        )                    −      y              (        1        −                              y            2                    3                +                              y            4                    5                −        .        .        .        .        )              =      lim          y      →              0                  1          −      1        =  −  1.We will have, that      lim          x      →              1                        ln      ⁡      (      x      )              arctan      ⁡      (      1      −      x      )        =  −  1.

What is the limit of the following function, as x approaches 1? <munder> <mo movablelimits="

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