Why $\lim_{x->0}log(x^x)=0$ ?. This means $0^0=1$?
For WolframAlpha $\lim_{x->0}log(x^x)=0$ but $0^0$ is not defined.
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In some calculations $0^0 = 1$. However that does not seem logical, because for example:
- $2^{-2} = 0.25; 0^{-2} = 0$
- $2^{-1} = 0.5; 0^{-1} = 0$
- $2^0 = 2^{-1} * 2 = 1; 0^0 = 0^{-1} * 0 = 0 ?!?!$
From these calculations we "know" that $0^0 = 0$. But $x \in \mathbb{R} \implies x^0 = 1$ ! And that's why the $0^0$ is undefined: because it has more than one value and breaks algebra with $0 = 1$...
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1This is not correct, as $0^{-1}={1\over 0}$ is not equal to $0$. Similarly for $0^{-2}={1\over 0^2}$, which is also undefined for the same reason (division by zero). – Per Erik Manne May 24 '14 at 07:24
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1No, only for $x>0$. The reason is that you get trouble with $a^{-b}={1\over a^b}$ otherwise. – Per Erik Manne May 24 '14 at 08:08
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@ciuak, It 'seems' to have more than one value, and thus is regarded 'indeterminate' – Shubham May 24 '14 at 09:26
$lon(x^x)=xlogx$.
$0^0$ is not defined,
$lim_{x\rightarrow0}log(x^x)=0$ is because of that $f(x)=x$ converges faster than $g(x)=\frac{1}{log(x)}$ at the point (0,0).
As we knew $\frac{1}{log(x)}log(x)=1$,so,when $x\rightarrow0$, $f(x)g(x)=o(1)$. That is to say $lim_{x\rightarrow0}log(x^x)=0$. – user152441 May 24 '14 at 06:40