1. Feb 8, 2007

### Born2Perform

all i write is in R U {infinite}

i understood that inf-inf is an indeterminate form, i don't understand why inf+inf is not. for example, doing all possible steps:

$$inf+inf = \frac{n}{0}+\frac{n}{0} = \frac{n*\frac{0*0}{0} + n*\frac{0*0}{0}}{0*0} = indeterminate$$

probably what i wrote is a non sense but can i have a proof that inf+inf=inf?

Last edited: Feb 8, 2007
2. Feb 8, 2007

### Mr.4

Hey n/0 can't be taken as infinite in math, or can it?! In physics problems most of the time n/0 would mean infinity but a lot less in math.

3. Feb 8, 2007

### Born2Perform

in math yes, in R no, for this reason i premitted that my set was R U {inf}

4. Feb 8, 2007

### arildno

You can't add non-numbers like n/0.

5. Feb 8, 2007

### Born2Perform

i imegined an answer like this, here ppl are also very restrictive about those
n/0 operations i don't understand why, i also made precise that i was writing in extended reals, i also did all possible steps without semplifying anything...

let's assume i have to solve a limit like

$$\lim_{\substack{x\rightarrow \infty\\y\rightarrow \infty}} x+y$$

and the result is obviously infinite, but why? i can see it only in the graphic?
what i wish to have is only a sentence that says "inf+inf is not indeterminate because bla bla...inf-inf is indeterminate because..." is there a criterion for which a thing is indeterminate?

6. Feb 8, 2007

### matt grime

Yes: if it can't be determined. The notation that some limit is infinty has explicit meaning. And under that meaning it is clear that the limit as x and y both tend to infinity of x+y is infinity again. However, it is not clear 'infinity -infinity' means. Consider 1/x and 1/x -k. As x tends to infinity, then either of those expressions tends to infinity as well. But 1/x - (1/x - k) =k, for any constant k.

7. Feb 8, 2007

### HallsofIvy

Staff Emeritus
Getting back to the original question- which has nothing to do with adding "infinity"!

It is possible to have examples two limits, each unbounded above (I am avoiding saying "going to infinity") such that the limit formed by subtracting the two converges to any desired quantity. Hence "indeterminant"- knowing that each does not converge tells you nothing about their difference.

If you have two limits, each unbounded above, then their sum must be unbounded and so does not converge- not "indeterminant" at all.

8. Feb 10, 2007

### ecurbian

In intuitive terms, the difference between two very large positive numbers can be small, even zero, thus indeterminate, but the sum of two very large positive numbers must be a very large positive number. Obviously (-inf)+(+inf)
is also indeterminate.

From the non-standard reals viewpoint, if d is infinitely small (a/d + b/d) = (a+b)/d, if a and b are standard, then (a+b) could be 0, making the result 0, which is not infinite, otherwise the result is infinite, but could be +inf or -inf. but if a and b are infinitesimal of the same order as d, then the value could be any standard value.

From the standard view point, if a,b, and d are functions of t, with limit as t goes to +inf of d being 0, while a and b remain positive, then the limit of the ratio (a+b)/d depends on the behaviour of (a+b) as t goes to infinity, and that could be almost anything.