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The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0 It is also an indefinite form because $$\infty^0 = \exp (0\log \infty) $$ but $\log\infty=\infty$, so the argument of the exponential is the indeterminate form zero times infinity discussed at the beginning. I'm perplexed as to why i have to account for this condition in my factorial function (trying to learn haskell).

Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number It says infinity to the zeroth power It seems as though formerly $0$ was considered i.

I heartily disagree with your first sentence

There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also gadi's answer) For all this, $0^0=1$ is extremely convenient, and i wouldn't know how to do without it In my lectures, i always tell my students that whatever their teachers said in school about $0^0$ being undefined, we. The intention is if you have a number whose magnitude is so small it underflows the exponent, you have no choice but to call the magnitude zero, but you can still salvage the.

0i = 0 0 i = 0 is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention 0x = 0 0 x = 0 On the other hand, 0−1 = 0 0 1 = 0 is clearly false (well, almost —see the discussion on goblin's answer), and 00 = 0 0 0 = 0 is questionable, so this convention could be unwise when x x is not a positive real. I began by assuming that $\dfrac00$ does equal $1$ and then was eventually able to deduce that, based upon my assumption (which as we know was false) $0=1$ As this is clearly false and if all the steps in my proof were logically valid, the conclusion then is that my only assumption (that $\dfrac00=1$) must be false.

In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$)

This is a pretty reasonable way to think about why it is that $0/0$ is indeterminate and $1/0$ is not However, as algebraic expressions, neither is defined Division requires multiplying by a multiplicative inverse, and $0$ doesn't have one. Defining 0^0 as lim x^x is an arbitrary choice

There are unavoidable discontinuities in f (x,y) = x^y around (0,0). Your title says something else than infinity times zero

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