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

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In summary, the conversation discusses finding the limit of x^((x^x)-1) as x approaches 0, which is an indeterminate form. The attempt at solving it involves using L'hopital's rule, but it is not a practical solution. An alternative approach is suggested, but it ultimately leads to the same limit and L'hopital's rule is still needed.
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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?
 
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  • #2
Suppose you let y = x^((x^x)-1), what is ln(y) = ?
 
  • #3
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?
 
  • #4
Oh whoops, I read too quickly. Yeah I don't see how you would do this without LH otherwise.
 

1. What is the limit of x^((x^x)-1) as x approaches 0?

The limit of x^((x^x)-1) as x approaches 0 is equal to 1.

2. How do you find the limit of x^((x^x)-1) as x approaches 0?

To find the limit of x^((x^x)-1) as x approaches 0, you can use l'Hopital's rule or evaluate the function at values closer and closer to 0.

3. What is the graph of x^((x^x)-1) as x approaches 0?

The graph of x^((x^x)-1) as x approaches 0 is a vertical line at x=0 with a y-value of 1.

4. Does the limit of x^((x^x)-1) as x approaches 0 exist?

Yes, the limit of x^((x^x)-1) as x approaches 0 exists and is equal to 1.

5. What is the significance of the limit of x^((x^x)-1) as x approaches 0 in mathematics?

The limit of x^((x^x)-1) as x approaches 0 is important in mathematics as it helps to understand the behavior of exponential functions as they approach 0. It also has applications in calculus and real analysis.

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