Infinity divided by itself sometimes? =0

[breezer]

I seem to remeber coming across this somewhere in college (10 years ago) in Calculus. I am a high school math teacher and would like to know if infinity divided by itself equals zero. If so, when and why? Under what circumstances? I do know that it is supposed to equal one.

[Tormod]

Not being a math wizard, I would assume that we would need to know what kind of infinity we are talking about.

[tom]

Infinity isn't a number so you can't use math operators on it. If you're talking about a limit , that's something different. The limit could be 0, 1 , or any other number ...

infinity / infinity is a case of indetermination.

 

lim x ^ (n-1) / x ^ n then x goes to infinity

 

is equal to infinity / infinity. When you simplify with x ^ (n-1)

 

lim x ^ (n-1) / x ^ n = lim 1 / x which is 0

 

lim x ^ n / x ^ n = lim 1 =1

 

lim 2 * x / x = 2

 

lim x ^ 2 / x = infinity

 

I know this aren't the best examples , but I hope it's helpful.

[maddog]

I seem to remeber coming across this somewhere in college (10 years ago) in Calculus. I am a high school math teacher and would like to know if infinity divided by itself equals zero. If so, when and why? Under what circumstances? I do know that it is supposed to equal one.

Tom is basically right. The representation of infinity divided by infinity is not a number per se. However,

you can take the ratio of two divergent series and if the denominator diverses at a less rate than the

numerator you could have a converging ratio in the limit.

 

Tom's examples were great.

 

To test for convergence you use L'Hospital's Rule (I learned in second semester Calculus) by taking the

derivative separately of both numerator and denominator. If the derivatives converge then the ratio

will converge. Neat huh... ;)

 

Maddog

[tom]

To test for convergence you use L'Hospital's Rule

 

or you could show it's Cauchy sequence , or use this ( I don't know it's name in English)

 

If |Xn - X |<= an for every n and an -> 0 then xn -> x

[Tim_Lou]

the graph of x/x is simply a line of y=1, so, graphically, it approaches to 1 as x approaches infinity.

 

well, tom is right, it depends on the increasing rate of the denominator and numerator....

if the information given is simply infinity over infinity, i would say it is undefined.

[Tim_Lou]

oh wait, you say "infinity divided by itself"

which means it is x/x

hehe, using L'Hospital's Rule:

lim x--> infinity (x/x) = lim x-->infinity (1/1) = 1

[tom]

Infinity divided my itself doesn't mean x/x

is could mean 2 * x / x or x+1 /x or x ^ 2 / x

[Tim_Lou]

by itself, it implies that the "infinity" and the "thing" that divides it are identical, therefore it is x/x

 

lim x--> infinity 2x / x can be infinity divided by infinity, but it is definitely not something divided by itself, since 2x and x approaches infinity differently.

[little cloud]

Isn't anything divided by itself equal to 1? Would that mean that two divided by two is the same as infinity divided by itself? ;)

[fusion]

I guess it might equal zero when x/x, (x being infinity) when the denominator is a infinately higher infinity than the numerator x. Lim x->infinity x/x^x=0 ;)

[maddog]

What I was saying "infinity divided by itself" is indefinite by itself. What I was implying

is what matter is the rate of divergence of the numerator over the denominator. If the

top diverges faster then the ratio diverges. That would be why L'Hospitals rule is good.

Lots' o' ways to get to infinity. The integers are countably infinite. The reals are

uncountably infinite which more infinite than the integers. One infinity (of some kind)

divided by another (of some or another kind) is not determinable until you know what

function you are taking the limit of.

 

lim (x -> inf) {x / x} = 1 because x/x == 1 for all x. I assumed this is not what was being

asked. What are you asking ? ;)

 

Maddog

[TINNY]

when you're talking about limits, i thought when you have an indeterminate form , you have to factorize the numerator and denumerator, or divide the numerator and denumerator by the variable with the highest power and see from there... if not, then the limit does not exist.

 

go here:

http://www.tpub.com/math2/36.htm

 

 

(it's been a long time since i jumped into any maths thread...)

[sanctus]

It is, as somebody said before, that infinity is not a number it is a concept and/or limit. Therefore if something isn't a number why should it equal to one, if divided by itself ? lets denote infinity as oo:

 

let's suppose oo is a number then if you add one to it you would obtain oo+1=oo' (the prime means just another infinity). There you see the paradox, if you add something to oo it stays oo by the definition of oo, but if it were a number it would have to be different. Therefore oo is not a number.

Now, you will agree that oo+oo=oo*2=oo, therefore 1=oo/oo=oo*2/oo=(oo/oo)*(2/oo)=1*0=0, therefore 1=0! Now you see why infinity divided by itself cannot be always 1, it all depends on how infinty is reached see post number 3

[Turtle]

I agree the original question is invalid because "infinity" is not a number as has been pointed out. Infinity simply means growth without bound,(which actually means no limit), which simply means you can always add one more. Since it's not a number, it is not succeptible to number operations in the way the question suggests. :)