Is it true that $0^r =0$ For all $r > 0$?
It comes in the context of real exponentiation as most of the text (e.g. Tao Analysis 1) defines $x^y$ for $x>0$ and not for $x=0$ when y is arbitrary real number
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Yes, this is the usual way to define $0^r,r>0$. – M. Winter Nov 01 '17 at 09:24
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This is a related question. It might be useful to take a look. – drhab Nov 01 '17 at 09:24
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It is easy to show from the very definition of exponentiation that $0^q=0$ holds for all rationals. – Nov 01 '17 at 09:27
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1@YvesDaoust Even negative rationals? – Frank Vel Nov 01 '17 at 09:29
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1@FrankVel: you are right to correct me. All positive rationals. (That was too implicit in my mind.) – Nov 01 '17 at 09:33
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Is it possible to show that $\lim_{n\to\infty}0^{a_n}=0^r$ where $a_n$ are rationals such that $a_n\to0$ – Jave Nov 01 '17 at 09:38
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1https://math.stackexchange.com/q/2499331/306553 didn't you just askthe question here? – Siong Thye Goh Nov 01 '17 at 09:40