use what god gave you and solve this

[absolute_deviation]

The problem is limits. You know questions like what is the limit on 1/n as n goes off to infinity? Of course in this case it is pretty simple that the limit is 0 as the function gets really close to zero but never actually becomes zero. another example is n^2/n. IN this case there is no limit (the limit is infinity). another example is 1^n. In this case the limit is just one. I was doing a bunch of these and i noticed a trend that made me think this:

When the degree of n in the numerator is higher than the degree of n in the denominator, the limit is infinity (in other words a limit does not exist). Can somebody please validate the preceding statement and provide a formal proof that either supports or doesn't support it.

[Kain]

I know how to solve this, go to your dads room, get his revolver from his desk [make sure its fully loaded] and place the barrel in your mouth, then slowly pull the trigger.

As to your question, i canot confirm, or deny your claim.

[Darz]

Originally posted by Kain
I know how to solve this, go to your dads room, get his revolver from his desk [make sure its fully loaded] and place the barrel in your mouth, then slowly pull the trigger.


I guess you can ask God himself on how to solve the problem if you do what Kain suggests...or maybe you'll be askin' SATanic Red man with horns to help you out. Or maybe you'll be lucky and you'll end up where all the dead pretty naked ladies hang out.

[Docta Beeeer]

Originally posted by Darz
I guess you can ask God himself on how to solve the problem if you do what Kain suggests...or maybe you'll be askin' SATanic Red man with horns to help you out. Or maybe you'll be lucky and you'll end up where all the dead pretty naked ladies hang out.
Um.

[Darz]

Originally posted by Docta Beeeer
Um.

I have problems...Ok?

[Downfall]

if you didnt we would worry

[Hobbes874]

Your trying to get us to do your math HW arn't you?

[Bhazard]

There is no such thing as infinity. Everything that has a beginning has an end.

[karny]

Originally posted by Bhazard
There is no such thing as infinity. Everything that has a beginning has an end.

thats bull shit, infinity is a concept and the consept is a real thing.

There may be nothing phisical that is infinite (that our minds can comprehend anyway, well our universe is suposed to be a closed infinite universe :D)

[absolute_deviation]

Originally posted by Hobbes874
Your trying to get us to do your math HW arn't you?

no man, i noticed this in class and i told my friend. She fucking told the teacher and now the teacher wants a proof from me.

If you guys could stop the arguement about infinity and help me out here.

[Darz]

Try posting the question on a math forum, you might get more help that way. :)

[absolute_deviation]

Originally posted by Darz
Try posting the question on a math forum, you might get more help that way. :)

tried googling but all the forums i found were deserted.

[Darz]

http://www.freemathhelp.com/forum/index.php


BAM! There ya go. It looks active too.

[Yian]

Originally posted by absolute_deviation
The problem is limits. You know questions like what is the limit on 1/n as n goes off to infinity? Of course in this case it is pretty simple that the limit is 0 as the function gets really close to zero but never actually becomes zero. another example is n^2/n. IN this case there is no limit (the limit is infinity). another example is 1^n. In this case the limit is just one. I was doing a bunch of these and i noticed a trend that made me think this:

When the degree of n in the numerator is higher than the degree of n in the denominator, the limit is infinity (in other words a limit does not exist). Can somebody please validate the preceding statement and provide a formal proof that either supports or doesn't support it.
Absolute, you need to elaborate more on your statement. Give some specs to the numerator and denominator. Here is what I found.

In usual cases, where D (denominator) and N (numerator) are both natural numbers (positive integer) but not zero, and when bigger than 1, then m (numerator degree) > n (denomenator) results infinity on its limit, because every set of m/n will be >1 and it approaches infinity as always, for there is no "largerst" positive integer.

But when N equals zero, the limit will be zero, D cannot be zero, which you already know. So this is one case.

Now let's see if it stands true when m<0, or n <0. Depending on the value of D and N, the result might vary, so here are the sets:

(X^(-m))/(Y^n), X > Y

(X^(-m))/(Y^n), X < Y

(X^m)(Y^(-n)), X > Y

(X^m)(Y^(-n)), X < Y

Notice how there is no determined rules on the final result of these 4 set will be greater than 1, or smaller than 1. This makes a big difference, because if N/D > 1, it is infinity, otherwise it approaches 1.

So, to reinforce your idea, I think this is more accurate:

When l(N^m)/(D^n)l =Q , m X n>1, and m>n, then Limit of Q = Infinity.

Notice that N x D does not have to be >1, because either +Q or -Q does not impact the direction of the number line. I haven't think about this last part thoroughly, so I could be wrong. But the way I see it, the absolute value took care of everything. If anyone thinks that the absolute value was not necessary for the limit of Q to be infinity, please correct me.

[absolute_deviation]

YIan man, thanks for all the hard work but you lost me.

[Yian]

:(

Which part don't you understand...??

[absolute_deviation]

(X^(-m))/(Y^n), X > Y

why do you have alternating signs on the exponents?

n is going off to infinity land in one direction.

[Yian]

Sorry, that's just a habbit of mine so when I do math I can always see them as a number, instead of getting distracted by the signs.

[absolute_deviation]

Originally posted by Yian
Sorry, that's just a habbit of mine so when I do math I can always see them as a number, instead of getting distracted by the signs.

so my idea is correct, right?

[WhoGivesARatsAss]

Your idea is sort of 80% correct, but what Yian said holds true and by infinity I assume that you meant both positive and negative infinity (both are treated the same in maths most of the time).

Yian solved the whole damn thing right there for you!

[cr3am]

Originally posted by Yian
But when N equals zero, the limit will be zero, D cannot be zero, which you already know. So this is one case.Sorry, fuzzy math, anything raised to 0 is 1. Nothing (besides 0) raised to any power is 0.
Wait, does he mean the numerator and denominator in the exponent? If so, man, you gotta specify...

[REBEL_AU]

The answer is 14.

[MrBored]

Erm, not sure what you were trying to explain, need to use word equation editor or something ;p but lemme try.

For the case lim(->oo) (n^2/n^1) this would tend towards infinity as the exponents cancel to leave n^1 or whatever, incase thats what you meant...

or lim(->oo) (x^n+1)/(y^(n)) is this what you meant? The degree of the numerator is higher than the denominator? This would go to (x/y) as the n's would cancel.

Erk, need equation editor or something.

[cr3am]

Originally posted by MrBored
or lim(->oo) (x^n+1)/(y^(n)) is this what you meant? The degree of the numerator is higher than the denominator? This would go to (x/y) as the n's would cancel. No they wouldn't.

[WhoGivesARatsAss]

Originally posted by cr3am
No they wouldn't.

Exacto-mundo-dudo-bitcho!

[Rohan]

Damn, I don't understand squat. All I see is a bunch of 321864f3s2d1fw86ef1a3s21f bersss.

[MrBored]

Originally posted by cr3am
No they wouldn't.
Hahahahaha, I just read what I wrote. I changed around what I was going to say and fucked it up horribly. What I was trying to say was that if you did long division ( with the numerator being a higher order polynomial ) you would get it down to a nicer expression to evalutate the limit. This should prove help prove what you're trying to say. But Absolute_dev probably has got bored with this now, so yeah...

[WhoGivesARatsAss]

You people make me sick!