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January 17th, 2018, 04:37 AM   #11
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Essentially your argument is one of context and the OP implies through the usr of R that we are talking about real numbers, where $0^0$ is undefined.

In other contexts different definitions or conventions do apply.

The example of $\displaystyle e^x=\sum_{n=0}\frac{x^n}{n!}$ is a rather amusing one. It's an expression packed with conventions and notational obfuscation.
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January 17th, 2018, 09:38 AM   #12
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Essentially your argument is one of context and the OP implies through the usr of R that we are talking about real numbers, where $0^0$ is undefined.

In other contexts different definitions or conventions do apply.
I never read anywhere in the OP that he's talking about the real numbers only. Sure $R$ can be a real number, it's the exponent that must be natural.

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The example of $\displaystyle e^x=\sum_{n=0}\frac{x^n}{n!}$ is a rather amusing one. It's an expression packed with conventions and notational obfuscation.
But all of math is packed with convention and notational obfucation!!!! Some like to define math as that which is invariant under notation.
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January 17th, 2018, 10:16 AM   #13
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Hmmmmm.... Mathematica does not like this one. See here.

-Dan
If you plot $\left(\dfrac 1 x\right)^{\frac 1 x}$

from 1 to some big number the graph certainly tends towards 1.
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January 17th, 2018, 11:20 AM   #14
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$\displaystyle \left( \frac1x \right)^{\frac1{\ln x}}$ doesn't
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January 17th, 2018, 11:25 AM   #15
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$\mathcal{O}(x)\gg\mathcal{O}(\log x)$

Last edited by greg1313; January 18th, 2018 at 07:53 AM.
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January 17th, 2018, 01:44 PM   #16
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Hmmmmm.... Mathematica does not like this one.
One can just do this or this.
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January 17th, 2018, 01:47 PM   #17
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$\mathcal{O}(x)\gg\mathcal{O}(\log x)$
Yes. More to the point $$x^{-\frac1{\ln x}}=e^{\ln{x^{-\frac1{\ln x}}}}=e^{-\frac1{\ln x}\ln x}=e^{-1}$$

Last edited by greg1313; January 18th, 2018 at 07:53 AM.
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