L'Hôpital's rule exercise with natural log functionI'm looking for some advice on the following exercise: lim x → 0 + ln ⁡ ( 1 x ) x This is my work so far: lim x → 0 + ln ⁡ ( 1 x ) x = lim x → 0 + x ⋅ ln ⁡ ( 1 x ) = lim x → 0 + ln ⁡ 1 x 1 x Taking the derivatives of the top and bottom: d d x ln ⁡ 1 x = − 1 x and d d x 1 x = − 1 x 2 So: lim x → 0 + − 1 x − 1 x 2 = lim x → 0 + x 2 x = lim x → 0 + x = 0Which doesn't quite add up. I'm thinking, I should rewrite the equation in the form of e x as the limit would then approach 1, which is the correct answer. However, I'm a bit stuck on that front.Any suggestions would be greatly appreciated!

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L&#039;Hôpital&#039;s rule exercise with natural log functionI&#039;m looking for some advice on the following exercise:      lim          x      →              0        +                        ln      ⁡              (                  1          x                    )        x  This is my work so far:      lim          x      →              0        +                        ln      ⁡              (                  1          x                    )        x    =      lim          x      →              0        +                  x    ⋅    ln    ⁡          (              1        x              )    =      lim          x      →              0        +                                ln        ⁡                              1            x                                      1        x            Taking the derivatives of the top and bottom:      d          d      x        ln  ⁡            1      x        =  −      1    x        and          d          d      x            1    x    =  −      1          x      2      So:      lim          x      →              0        +                                −                  1          x                            −                  1                      x            2                                =      lim          x      →              0        +                                x        2            x        =      lim          x      →              0        +                  x    =  0Which doesn&#039;t quite add up. I&#039;m thinking, I should rewrite the equation in the form of       e    x   as the limit would then approach   1, which is the correct answer. However, I&#039;m a bit stuck on that front.Any suggestions would be greatly appreciated!

Abigayle Dominguez · Accepted Answer

lim          x      →      0        ln  ⁡            (              1        x            )        x    ≠      lim          x      →      0        x  ln  ⁡      (          1      x        )  You have that  ln  ⁡            (              1        x            )        x    =      e          x      ln      ⁡      (              1        x            )        .Since      lim          x      →      0        x  ln  ⁡      (          1      x        )    =      lim          x      →      0                  ln      ⁡              (                  1          x                )                    1      x            =    (Hop.)        lim          x      →      0                                    −          1                          x          1                            1        x                    −              1                  x          2                      =      lim          x      →      0        x  =  0and that   x  ↦      e    x   is continuous at   x  =  0, we can conclude that      lim          x      →      0        ln  ⁡            (              1        x            )        x    =      e    0    =  1.

chiodaiokaC · Accepted Answer

One should be careful about the exponent, there are two possible meanings:  ln  ⁡      (                  (                  1          x                )                    x              )    ≠            (      ln      ⁡              (                  1          x                )            )              x      Let us compute the limit of both of them.      lim          x      →              0                  +                      ln  ⁡      (                  (                  1          x                )                    x              )    =      lim          x      →              0                  +                      x  ln  ⁡      (          1      x        )    =      lim          x      →              0                  +                      x      (    ln    ⁡    1    −    ln    ⁡    x    )    =  −      lim          x      →              0                  +                      x  ln  ⁡  x  =  0.However,                              lim                      x            →                          0                              +                                                                          (            ln            ⁡                          (                              1                x                            )                        )                                x                                              =                              lim                      x            →                          0                              +                                                              e                      ln            ⁡                                          (                ln                ⁡                                  (                                      1                    x                                    )                                )                                            x                                                                                      =                              lim                      x            →                          0                              +                                                              e                      x            ln            ⁡                          (              ln              ⁡                              (                                  1                  x                                )                            )                                                                        =                              lim                      x            →                          0                              +                                                              e                                                    ln                ⁡                                  (                  ln                  ⁡                                      (                                          1                      x                                        )                                    )                                                            (                                  1                  x                                )                                                                                      =                              lim                      x            →                          0                              +                                                              e                                                                      [                  ln                  ⁡                                      (                    ln                    ⁡                                          (                                              1                        x                                            )                                        )                                    ]                                                  ′                                                                              (                                      1                    x                                    )                                                  ′                                                                                                      =                              lim                      x            →                          0                              +                                                              e                                                                                          (                    ln                    ⁡                                          (                                              1                        x                                            )                                        )                                                        ′                                                                    (                  ln                  ⁡                                      (                                          1                      x                                        )                                    )                                                            −                                  1                                      x                                          2                                                                                                                                            =                              e                      −                          lim                              x                →                                  0                                      +                                                                                                      (                −                                  1                  x                                )                                            ln                ⁡                                  (                                      1                    x                                    )                                                                    x                              2                                                                                      =                              e                      −                          lim                              x                →                                  0                                      +                                                                                      x                              ln                ⁡                x                                                                                      =                              e                      0                          =        1.

L'Hôpital's rule exercise with natural log function I'm looking for some advice on the following exercise: lim_(x -> 0^+)(ln 1/x)^x

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