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# Limit of $x^x$ as x tends to 0. watch

1. I have been trying to work this out. It's confusing because wouldn't it just be , and anything raised to the power of 0 is 1 but 0 raised to any power is 0?

I rewrote , but now I am not really sure what to do as when , .
2. e^xlnx = e^((lnx)/(1/x))
3. (Original post by NotNotBatman)
e^xlnx = e^((lnx)/(1/x))
I don't see how that helps: ln x/1/x. Now if x-->0 then the numerator will tend to negative infinity and the denominator will tend to positive infinity. Is such a quotient defined?
4. (Original post by FXLander)
I don't see how that helps: ln x/1/x. Now if x-->0 then the numerator will tend to negative infinity and the denominator will tend to positive infinity. Is such a quotient defined?
You can apply L’Hopital’s on the quotient
5. (Original post by FXLander)
I have been trying to work this out. It's confusing because wouldn't it just be , and anything raised to the power of 0 is 1 but 0 raised to any power is 0?

I rewrote , but now I am not really sure what to do as when , .
I think that you could look at . It's easy to show that this is well-defined, decreasing and negative for so as , we can consider it as an increasing function with supremum 0 (or alternatively look at its reflection ). So what is ?
6. (Original post by atsruser)
I think that you could look at . It's easy to show that this is well-defined, decreasing and negative for so as , we can consider it as an increasing function with supremum 0 (or alternatively look at its reflection ). So what is ?
It is 0. How do you best show that it is well-defined, decreasing and negative for 0<x<1/e. I was thinking of finding f'(x) = 1+ ln x, but not really sure how to continue from that.

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