I hope you don't put as a duplicate my question as it is thought for the specific case I am going to show you now, which is about a calculation.

I read about the topic of infinity over infinity, which is considered to be undefined, and I read also this thread Is infinity a number? which is interesting, although more historical and philosophical, thus less relevant for the question I ask.

If I have to find the limit as x approaches infinity for the rational function $(x^2-x-2)/(x^2-1)$ , I would proceed in the following way (just for coherence, because in a previous exercise with the same function, but with the limit which approaches 1, I made the same algebraic steps):

lim x which approaches infinity, $(x-2)(x+1)/(x-1)(x+1)$ , applying the properties of limits and an algebraic property: lim x which approaches infinity $(x-2)$ * lim x which approaches infinity $(1/x-1)$ , plugging: (infinity - 2 = infinity) * (1 / infinity - 1 = infinity; 1 / infinity = 0).

Thus: infinity * 0 = undef.

First question: is it a convention that infinity * zero and zero * infinity are undefined? Could be applied the commutative property in this case? Can't we interpret in any way this operation with 1 as a result?.

Also if I make the same calculation in this other way (although, perhaps, incorrect from an algebraic point of view...I am not sure, actually, about this...as I don't resolve the limit process inside the parenthesis) I end up with infinity * (1/infinity) = infinity / infinity and the result should be interpreted as undefined. The same is taking the limit at minus infinity, I end up with -infinity * (1 / - inifnity) = -infinity / - infinity = undefined.

However, I know from the graph that taking the limit at infinity to this function gives exactly 1! Does this mean that saying that the operations are undefined is an arbitrary fact or I am making mistakes in my algebraic steps?.

If I put the function in a calculator, this choose different algebraic steps - applying more than one algebraic property - for avoiding the kind of situation I have showed, namely, infinity over infinity, minus infinity over minus infinity, 0 * infinity, as in the calculator we evaluate finally: $(1-2/x) / (1-1/x)$, which gives $1-0 / 1-0 = 1$.

However, since that I don't see clear algebraic mistakes in my passages, I ask you this question. From the reading I am having, I understand that there is debate upon infinity over infinity and it isn't so obvious considering the operations showed above as undefined and, perhaprs, the choice depends on the branch of mathematics and the operation we are performing or am I totally wrong?

More practically, do you suggest me to use the passages of calculator for my exercise?

PS. This is the output of the calculator but, please, have a look at my passages too https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Cleft%28x%5E%7B2%7D-x-2%5Cright%29%7D%7Bx%5E%7B2%7D-1%7D/?origin=button

  • $\begingroup$ Ok, reading again my question, I agree that is indeed intuitive considering infinity multiplied zero and the revers to be undefined. But where I'm wrong in my calculation? $\endgroup$ – Always learning Nov 5 '15 at 10:08
  • $\begingroup$ I know that I have written long but the question is easy. Please, answer! I've spent much time writing this! $\endgroup$ – Always learning Nov 5 '15 at 10:17
  • $\begingroup$ In some parts of analysis, notably measure theory, 0 times infinity is defined to be zero. $\endgroup$ – yoyostein Nov 5 '15 at 15:14

Not sure I understand your question entirely but here my answer.

Infinity is, as you know, not a number, and thus we can't use it in an expression. There is no such thing as $0\cdot\infty$, nor is there $\frac{\infty}{\infty}$.

HOWEVER, we can define this with limits. The first one would be the limit if $xy$ as $x$ goes to $0$ and $y$ goes to infinity.

This is called "indeterminate", not undefined. Indeterminate means that we do not yet know what the value of the expression is. It could be $0$, $1$, $42$, or go off to infinity.

The limit if $\frac{x}{y}$ as both $x$ and $y$ go to infinity is the same thing. It could be anything, since there are many 'types' of infinity.

For example, $\frac{2x}{x}$ approaches $2$ as $x$ goes to infinity, but $\frac{x^2}{x}$ only gets bigger and bigger.

In your case, the expression approaches $1$ as $x$ goes to infinity. This is NOT the same thing as saying $\frac{\infty}{\infty}=1$, which is meaningless.

  • $\begingroup$ Thank you, could you show me how you would make the calculation with the function of my question, please? Perhaps, in this way, I can have my doubts cleared. $\endgroup$ – Always learning Nov 5 '15 at 10:38
  • $\begingroup$ I sayed that because making calculation I ended up with infinity over infinity and I know that the function approaches 1 as x goes to infinity $\endgroup$ – Always learning Nov 5 '15 at 12:31
  • $\begingroup$ In that case the function approaches 1. The reason, basically, is because both the numerator and denominator approach $x^2$ as x gets large $\endgroup$ – Elliot G Nov 5 '15 at 22:11

∞-∞ is undefined, however A=∞; A-A=0 works If the ∞s are different their relationship needs to be known from earlier principals.


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