Answer
Dear David
That way “playing with infinity” is “dangerous”. It is important to remember that infinity is not a real number, so the normal calculation rules for real numbers do not simply apply to infinity. We can add the elements “-∞” and “+∞” to the real numbers. Then for every real number x holds: -∞ < x < +∞.
We can extend a number of operations without any problems so that we can also ‘calculate’ with these new elements. Examples of this are (x is always a real number):
(+∞) + x = x + (+∞) = +∞
(-∞) + x = x + (-∞) = -∞
(+∞) + (+∞) = +∞
(-∞) + (-∞) = -∞
x*(+∞) = (+∞)*x = +∞ if x > 0
x*(+∞) = (+∞)*x = -∞ if x < 0
x/(+∞) = x/(-∞) = 0 if x ≠0
…
We can indeed put your example of 2*(+∞) = +∞ like this. But suppose at this point you were to “divide both members by +∞” to “delete +∞” on both sides, then it would say 2 = 1 and that’s obviously wrong. The rule that you may divide both members by a real number (other than 0) does not apply to “+∞” (or “-∞”).
In addition to the list of calculation rules above, there are also a number of expressions that we cannot assign a value to. Examples of this are:
(+∞) – (+∞), (+∞) + (-∞), 0*(±∞), 1^(+∞), (±∞)/(±∞), …
So your example of 0*(+∞) also belongs here. We cannot define these operations meaningfully if we want to keep all our old rules for calculating the real numbers.
The cases above where things “go wrong” are also called “indefinite forms”. We find the reason for this in (limits of) functions. Suppose there is a function with prescription f(x) going to 0 when x goes to 0 and there is a second function with prescription g(x) where g(x) goes to +∞ as x goes to 0. We can then see what the product f(x)*g(x) does. With just “fill in” you would get the form 0*(+∞), but what can happen?
a. f(x) = x² and g(x) = 1/x, then f(x)*g(x) = x²/x = x, this goes to 0 when x goes to 0,
b. f(x) = x and g(x) = 1/x, then f(x)*g(x) = x/x = 1, this remains 1 if x goes to 0,
c. f(x) = x and g(x) = 1/x³, then f(x)*g(x) = x/x³ = 1/x², this goes to +∞ as x goes to 0.
So we have found here three examples of the indefinite form 0*(+∞) for which the limit is always slightly different; 0, 1 and even +∞. By in case b. for g(x) = k/x taking k an arbitrary real number, you can even get any k you want as an outcome.
Because with 0*(+∞) you can get a different value each time, depending on the “way in which” you get that 0 and that +∞, we just call that expression an “indefinite form”. The same goes for the other indefinite forms in the previous list.
On the things you write down in the “reasoning”, a few comments:
– 0.999… is indeed 1, if you are talking about real numbers and that first is an infinitely continuous decimal development with all nines;
– 0.000…1 meaning “infinitely many zeros and then a 1” does not exist; if there are infinitely many zeros you can’t write another 1 “after that”, the number 0.000… is just 0;
– What is ‘correct’ (in fact: a meaningful definition) is 1/(+∞) = 0; then also in the first list; however, you may not conclude from this that 0*(+∞) = 1;
– 10X.10-X is indeed equal to 1, for every real x; the expression 10+∞.10-∞ however, is an indefinite form (see earlier); the “dussss” after that is therefore dangerous!
Calculating with infinity may be fun and interesting, but as you can see, you have to be very careful.
regards
Tom