Normally, $a^b$ is defined as $e^{b \ln(a)}$ and $e^x$ is defined by its Maclaurin Series, and $\ln(x)$ is defined as the real inverse of $e^x$. If we try this with $0^x$, we get $e^{x \ln(0)}$, which is obviously incorrect. Of course we can just define it separately, and say $0^x = 0$ for $x \gt 0$ (or $\geq$ 0 if you're feeling spicy), but this feels inelegant.
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2All mathematicians agree in $0^x=0$ for positive $x$. A lot mathematicians have very good reasons (the binomial theorem, Taylor expansion are only two of them) to define $0^0=1$. – Michael Hoppe Jul 04 '21 at 08:17
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Use the Maclurin series you mentioned to find that $$\lim_{\delta \to 0^+}\delta^x=0$$ and hence $0^x=0$ (of course for positive $x$ only).
Does that help?
Sayan Dutta
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Yes, but you're telling nothing new: we all agree in $0^x=0$ for positive $x$ ... – Michael Hoppe Jul 04 '21 at 09:13
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1@MichaelHoppe Yes, that was what the question asked, I guess. It asked why do we agree on that. And, I gave a reason to do so. – Sayan Dutta Jul 04 '21 at 09:14