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Bookmark or Share- Why is $\\infty\\times 0$ indeterminate? - Mathematics Stack ExchangeYour title says something else than "infinity times zero". It says "infinity to the zeroth power". 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.
- calculus - any number raised to the power of infinity - Mathematics ...$\begingroup$ i am assuming from your answer that if a>1 then limit x--> + infinity or - infinity of a to the power x is = 0 $\endgroup$ – Zia ur Rahman Commented Oct 5, 2011 at 13:07
- One divided by Infinity? - Mathematics Stack ExchangeSimilarly, the reals and the complex numbers each exclude infinity, so arithmetic isn't defined for it. You can extend those sets to include infinity - but then you have to extend the definition of the arithmetic operators, to cope with that extended set. And then, you need to start thinking about arithmetic differently.
- limit when zero divided by infinity - Mathematics Stack ExchangeI know that infinity is not a real number but I am not sure if the limit is indeterminate. (Also, there are people who are saying contradictory things on internet) I know very well that it is not possible to use Hopital's rule.
- What is infinity divided by infinity? - Mathematics Stack ExchangeEssentially, you gave the answer yourself: "infinity over infinity" is not defined just because it should be the result of limiting processes of different nature. I.e., since such a definition would be given for the sake of completeness and coherence with the fact "the limiting ratio is the ratio of the limits", your
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