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  1. Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$?

  2. 2 Φεβ 2020 · This is a falsidical paradox arising from incorrect reasoning about the nature of square roots .

  3. 28 Ιουλ 2023 · If we start only with \(a^1=a\) and the product rule, then we can immediately prove that \(a^0=1\) because \(a^0\cdot a=a^0\cdot a^1=a^{0+1}=a^1=a\), and dividing through by a (which is assumed not to be zero), we conclude that \(a^0=1\). But then for any positive integer n, $$a^n=a^{\overset{n\text{ times}}{\overbrace{1+1+\cdots+1}}}=\overset ...

  4. The error is that the "..." denotes an infinite sum, and such a thing does not exist in the algebraic sense. The usual way to make sense of adding infinitely many numbers is to use the notion of an infinite series: We define the sum of an infinite series to be the limit of the partial sums.

  5. 4 Δεκ 2010 · A clever bit with that proof of 0^0 = 2 is that along with 0^0 = 1, and 0^0 = 0, it actually lets you define it as any value you like. If you can do it for 2, you can do it for 3 and every other positive integer, using a bit of induction.

  6. Is this true or false? \ [ 0^0 = 1 \] Why some people say it's true: A base to the power of \ (0\) is \ (1\). Why some people say it's false: An exponent with the base of \ (0\) is \ (0\). Reveal the correct answer.

  7. 11 Αυγ 2023 · Why does 0! = 1? Is there a reason or is this like anything to the power of 0 = 1 - there is not a reason? Recall that “factorial” means the product of descending integers, and is written as \(n!=n(n-1)(n-2)\dots1\). For example, \(5!=5\cdot4\cdot3\cdot2\cdot1=120\).

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