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

1. Dec 6, 2008

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

2. Dec 6, 2008

mutton

Now x^x is the exponent and x is the base, so do exactly as you just explained.

3. Dec 6, 2008

emilkh

Could you elaborate more? I alredy tried this method and got stuck with
lim (ln x) * (x^x), i could not solve it

4. Dec 6, 2008

Hurkyl

Staff Emeritus
Why are you stuck? You already said you knew what to do with x^x....

5. Dec 6, 2008

emilkh

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)

6. Dec 6, 2008

mutton

Be careful; $$- \infty \cdot 0$$ is indeterminate.

Edit: I see, nevermind.

Last edited: Dec 6, 2008
7. Dec 6, 2008

emilkh

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

8. Dec 6, 2008

poutsos.A

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

9. Dec 6, 2008

emilkh

[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