1. Originally Posted by Jose27
The problem is, Wolfram Alpha is calculating the limit in the complex plane with the definition I first gave, but by what I understood the limit that was asked ignores this and uses only the fact negative numbers have odd roots (ie. we want the limit of a sequence, not the limit of the function)
Actually, the domain of $x$ isn't specified, so I'm a little confused as well. It's actually a question from Wade's Introduction to Analysis (4th Ed), and the answer given is 1. It does seem that it's the sequence it's referring to, then?

2. Originally Posted by h2osprey
Actually, the domain of $x$ isn't specified, so I'm a little confused as well. It's actually a question from Wade's Introduction to Analysis (4th Ed), and the answer given is 1. It does seem that it's the sequence it's referring to, then?
I don't know. If the book says the solution is 1 then it can't be the sequence since that one converges to -1, but otherwise it wouldn't make sense writing x ->0+. I'm confused.

3. Originally Posted by Jose27
I don't know. If the book says the solution is 1 then it can't be the sequence since that one converges to -1, but otherwise it wouldn't make sense writing x ->0+. I'm confused.
The book probably made the same mistake I did.

4. Originally Posted by mr fantastic
$(\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}$. So you should first consider $\lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ which has the indeterminant form $\frac{\infty}{\infty}$. Using l'Hopitals rule is an obvious thing to do.
Sorry, I'm not getting how you might get the answer 1 from this?

I thought you couldn't use L'Hopital's on this $\lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ as the numerator is undefined, and hence I don't see how you arrive at the indeterminate form $\frac{\infty}{\infty}$.

5. Originally Posted by h2osprey
Sorry, I'm not getting how you might get the answer 1 from this?

I thought you couldn't use L'Hopital's on this $\lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ as the numerator is undefined, and hence I don't see how you arrive at the indeterminate form $\frac{\infty}{\infty}$.
Read my previous post! I made a mistake (sheesh, if the whole world didn't know, it does now)!

6. Originally Posted by mr fantastic
Read my previous post! I made a mistake (sheesh, if the whole world didn't know, it does now)!
Ah, got it. Sorry for the confusion

Page 2 of 2 First 12