Limit of x x as x tends to 0I am trying to solve the following limit: lim x → 0 x x The only thing that comes to mind is to write x x as e x ln ⁡ x and getting the right sided limit would be easy but I don't see how I could get the left sided one seeing that the ln is not defined for negative numbers.Is there something I am missing or is there another way to go about it?P.S.:I don't know anything about derivatives so please keep it to the limits.

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Limit of       x    x  as x tends to 0I am trying to solve the following limit:      lim          x      →      0            x    x  The only thing that comes to mind is to write       x    x   as       e          x      ln      ⁡              x             and getting the right sided limit would be easy but I don&#039;t see how I could get the left sided one seeing that the   ln is not defined for negative numbers.Is there something I am missing or is there another way to go about it?P.S.:I don&#039;t know anything about derivatives so please keep it to the limits.

feaguelaBapzo · Accepted Answer

First find       lim          x      →      0        x  ln  ⁡  (  x  )  =      lim          x      →      0                  ln      ⁡      (      x      )              1              /            x      . Using L&#039;Hospital this become       lim          x      →      0                  1              /            x              −      1              /                    x                  2                      =      lim          x      →      0        −  x  =  0.So       lim          x      →      0            e          x      ln      ⁡      (      x      )        =      e                  lim                  x          →          0                    x      ln      ⁡      (      x      )        =  1.(note assuming   x  &amp;gt;  0 of course, since       x          x       is not well-defined otherwise)Also, if you allow   x  &amp;lt;  0 but   x must be rational only, then the limit do not exist. This can be seen from the fact that       lim          x      →      0            x          x        =  1 when   x  &amp;gt;  0. This means, that there are positive   x arbitrarily close to   1 in any neighbourhood of   0. Now once you look at negative   x, then by considering rationals with odd denominator in any neighbourhood of   0, then the once with even numerator will be positive, and odd numerator will be negative. The positive one will be arbitrary close to   1, the negative one will be arbitrarily close to   −  1, so there are no limit.If you allow   x  &amp;lt;  0 and   x must be rational only, but also allow only a subset of rational such that       x          x       have definite sign, then the limit is either   1 or   −  1 from the left.

deceptie3j · Accepted Answer

L&#039;Hospital&#039;s rule is quickest. I show that other approaches are possible:For      x  ∈      [               1      7        ,               1      6             )    ,            (              1        7            )                      1        6              &amp;lt;      x    x    &amp;lt;  1.Now using Newton&#039;s Binomial expansion,                                          (            1            −            α            )                                1            6                          =        1        +                  1          6                (        −        α        )        +                                            (                              1                6                            )                                      (              −                              5                6                            )                                            2            !                          (        −        α                  )          2                +                                            (                              1                6                            )                                      (              −                              5                6                            )                                      (              −                              11                6                            )                                            3            !                          (        −        α                  )          3                +        …                                          =        1        −                  1          6                α        −                              (                          1              6                        )                    2                                      (                          5              6                        )                    1                          α          2                −                              (                          1              6                        )                    3                                      (                          11              6                        )                    2                                      (                          5              6                        )                    1                          α          3                −        …                                                   &amp;gt;        1        −                  1          6                          (          α          +                      1            2                                α            2                    +                      1            3                                α            3                    +          …          )                                                  =        1        −                  1          6                (                 −        ln        ⁡        (        1        −        α        )                 )        =        1        +                  1          6                (                 ln        ⁡        (        1        −        α        )                 )        .                          Substituting      α  =      6    7    ,    into the above, we see that  1  +      1    6    ln  ⁡      (          1      7        )    =  1  −      1    6    ln  ⁡  (  7  )  &amp;lt;            (              1        7            )                      1        6              &amp;lt;      x    x    &amp;lt;  1    ∀  x  ∈      [               1      7        ,               1      6             )    .More generally, for any      k  ∈      N    ,    we have,  1  −            ln      ⁡      (      k      +      1      )        k    &amp;lt;            (              1                  k          +          1                    )                      1        k              &amp;lt;      x    x    &amp;lt;  1    ∀  x  ∈      [               1              k        +        1              ,               1      k             )    .Letting      k  →  ∞    yields the result.

Limit of x x </msup> as x tends to 0 I am trying to solve the following limit:

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