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

by emilkh
Tags: limit, x>0
 P: 7 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
 P: 179 Now x^x is the exponent and x is the base, so do exactly as you just explained.
 P: 7 Could you elaborate more? I alredy tried this method and got stuck with lim (ln x) * (x^x), i could not solve it
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P: 16,091
Limit of x^x^x as x->0 from right

 Quote by emilkh Could you elaborate more? I alredy tried this method and got stuck with lim (ln x) * (x^x), i could not solve it
Why are you stuck? You already said you knew what to do with x^x....
 P: 7 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)
P: 179
 Quote by emilkh lim ln x * (x^x) = lim ln x * lim x ^ x = (- infinity) * 0 = 0 as X -> 0+
Be careful; $$- \infty \cdot 0$$ is indeterminate.

Edit: I see, nevermind.
P: 7
 Quote by mutton Be careful; $$- \infty \cdot 0$$ is indeterminate.
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
P: 101
 Quote by emilkh 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
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
 P: 7 [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

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