What is the Limit of x^((x^x)-1) as x Approaches Zero?

  • Thread starter Thread starter tsuwal
  • Start date Start date
  • Tags Tags
    Limit
tsuwal
Messages
105
Reaction score
0

Homework Statement


Limit of x^((x^x)-1) as x->0


Homework Equations


Lim x^x=1
x->0


The Attempt at a Solution


it's an 0^0 indetermination so I tried to solve it the usual way, by first calculating the limit of log(x)*(x^x-1) as x->0 with L'hopital's rule. I got e^0=1 confirmed by Wolfram. However, the using L'hopitals rule on this limit is not very practical, is there a better soluction?
 
Physics news on Phys.org
Suppose you let y = x^((x^x)-1), what is ln(y) = ?
 
I would get the same limit log(x)*(x^x-1) as x->0 but solving this by L'hopital's rule takes a while or is it the fastest way?
 
Oh whoops, I read too quickly. Yeah I don't see how you would do this without LH otherwise.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
5
Views
1K
Replies
8
Views
2K
Replies
5
Views
2K
Replies
19
Views
2K
Replies
8
Views
1K
Replies
3
Views
4K
Replies
15
Views
6K
Back
Top