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

[emilkh]

Any ideas?
lim X^XX
x-> 0⁺

I know how to do x^x:
lim X^X = lim e^{x * ln x} = e⁰ = 0
lim x * ln x = lim ln x / (1/x) = lim (1/x) / (-1/x²) = lim -x = 0

 

[mutton]

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

 

[emilkh]

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

 

[Hurkyl]

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...

 

[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 e^{x^x *ln x} = e^0 = 1?

The limit suppose to be 0, .000001 ^ ( .000001 ^ .000001 ) = very very very small number (checked with calculator)

 

[mutton]

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.

 

[emilkh]

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 going to work

 

[poutsos.A]

Any ideas?
lim X^XX
x-> 0⁺

I know how to do x^x:
lim X^X = lim e^{x * 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

 

[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... got to take break