Limit of x^x^x as x->0 from right

  1. Any ideas?
    lim XXX
    x-> 0+

    I know how to do x^x:
    lim XX = lim ex * ln x = e0 = 0
    lim x * ln x = lim ln x / (1/x) = lim (1/x) / (-1/x2) = lim -x = 0
     
  2. jcsd
  3. Now x^x is the exponent and x is the base, so do exactly as you just explained.
     
  4. Could you elaborate more? I alredy tried this method and got stuck with
    lim (ln x) * (x^x), i could not solve it
     
  5. Hurkyl

    Hurkyl 16,089
    Staff Emeritus
    Science Advisor
    Gold Member

    Why are you stuck? You already said you knew what to do with x^x....
     
  6. lim ln x * (x^x) = lim ln x * lim x ^ x = (- infinity) * 0 = 0 as X -> 0+

    So you want to tell me that lim (X^x^x) = lim ex^x *ln x = e^0 = 1?

    The limit suppose to be 0, .000001 ^ ( .000001 ^ .000001 ) = very very very small number (checked with calculator)
     
  7. Be careful; [tex]- \infty \cdot 0[/tex] is indeterminate.

    Edit: I see, nevermind.
     
    Last edited: Dec 6, 2008
  8. Well great! This is where I am stuck.

    lim ln x * x was solved by switching it to lim ln x / (1/x) and taking derivatives, with lim ln x * x/x it's not gonna work
     
  9. e0 is not =0, but...........1.............. here is your mistake ,

    hence ......limx^x=1 as x tends to 0 from the right

    And lim (ln x) * lim x^x = infinity multiplied by 1 and NOT by 0
     
  10. [edited for content], in my solutions it was 1, but somehow I copied formula wrong and the whole time assumes lim x^x = 0. I spend way too much time studying for finals.... gotta take break
     
    Last edited by a moderator: Dec 6, 2008
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