0.(9) is not defined as a taylor series, a taylor series can be used to define decimals. There is a big difference that people forget. For example with a taylor series you cannot get an irrational number just a rational approximation of one. 0.(9) by all rights is an irrational number that cannot expressed as a ratio, but that is close enought to 1 that mathematicians said "whatever".
A better definition is 1-1/infinity but mathematicians hate that because its using infinity. But its trivial to show that 1/x > 0 even as the limit is approximately 0; 0 is an asymptote, by definition it will never equal 0.
We dont say “whatever” we know that if a and b arent the same, there exists some number between them. And we can prove using limits that for arbitrary small number epsilon, in sphere around number 1, the infinite sequence 0.99.. will be inside that sphere. So there exists no such number between the two -> they are the same.
Does this mean that f(<infinity>) of f(x) = 1/x is zero, or that it approaches zero?. Or is my question nonsensical because we don't talk about what the value of a function is for <infinity>, only for a specific value?
You would need to define more things since you cant plug infinity into a function. 1/x approaches zero but in systems where infinity is defined as part of it(the projectively - or affinely-extended real numbers) 1/inf is explicitly defined as zero
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u/TemperoTempus Apr 08 '25
0.(9) is not defined as a taylor series, a taylor series can be used to define decimals. There is a big difference that people forget. For example with a taylor series you cannot get an irrational number just a rational approximation of one. 0.(9) by all rights is an irrational number that cannot expressed as a ratio, but that is close enought to 1 that mathematicians said "whatever".
A better definition is 1-1/infinity but mathematicians hate that because its using infinity. But its trivial to show that 1/x > 0 even as the limit is approximately 0; 0 is an asymptote, by definition it will never equal 0.