I want to know Zero power to Zero equal to 1 or Indeterminable. I think it cannot be exist. Please explain with proper mathematical definitions.
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This is defined to be 1 for convenience. For example, in the Taylor series of $e^x = \sum_0^\infty x^n/n!$, you have $e^0=\sum_0^\infty 0^n/n! = 0^0/0!+0+0+...$. Since $0!=1$, you had better have $0^0$ also equal to $1$ in order to get the correct result. – Cass Apr 04 '14 at 05:40
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It depends from the context, $0^0=1$ however when taking a limit $f(x)^{g(x)}$ knowing that $f(x)\to 0$ and $g(x)\to 0$ doesn't imply that $f(x)^{g(x)}\to 1$ – Alessandro Codenotti Apr 04 '14 at 05:45
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@Alessandro It doesn't? I wasn't aware of that. Could you provide an example? ( I mean apart from the case where $f$ goes to $0$ "faster" than $g$) Is there anything else? – Guy Apr 04 '14 at 06:04
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Actually I was thinking of the trivial $\lim\limits_{x\to 0^+}0^x=0$ when I wrote my comment. According to wikipedia as long as $f(x)$ and $g(x)$ are analityc (at the interested point) and $f(x)>0$ in a neighbourhood of said point the limit is $1$ – Alessandro Codenotti Apr 04 '14 at 08:04