Author Topic: Math Help  (Read 1091 times)

Kuiper7986

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Math Help
« on: November 26, 2003, 12:47:35 am »
Is this formula true?

L\'H?pital\'s rule:

If is of the form , and if exists, then .
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seperot

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« Reply #1 on: November 26, 2003, 12:53:00 am »
the answers 3

dont ask how i know but trust me

its 3

Kuiper7986

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« Reply #2 on: November 26, 2003, 12:56:39 am »
uhh its not an equation, your not supposed to solve for anything, I\'m just asking if that statement is true or not, SMARTY PANTS :P
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Leander256

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« Reply #3 on: November 26, 2003, 12:58:51 am »
Quote
Originally posted by seperot
the answers 3

dont ask how i know but trust me

its 3


No way! The answer is 42!

seperot

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« Reply #4 on: November 26, 2003, 01:03:42 am »
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No way! The answer is 42!


of course the same as the answer to the meaning of life :P

CadRipper

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« Reply #5 on: November 26, 2003, 01:12:18 am »
Quote
Originally posted by Kuiper7986
Is this formula true?

L\'H?pital\'s rule:

If is of the form , and if exists, then .

True, of course, and it also applies to the form 0/0. Note the recursive possibility.

Xandria

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« Reply #6 on: November 26, 2003, 01:53:25 am »
Dangit, it\'s been less than a year since my calculus class, and I couldn\'t remember the answer to this!

My brain is rotting!!!!  8o

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TheGeneral

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« Reply #7 on: November 26, 2003, 06:35:32 am »
He is right you know, the Hospital\'s guy rule works for any form of 1/0 or 0/1, but not infinity/infinity, I dun think. Don\'t quote me on this.

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lynx_lupo

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« Reply #8 on: November 26, 2003, 08:37:10 am »
I\'ve never heard of that rule, but the way I see it, the first limit(?) is 1 and the second limit\'s parameters are just diferentials, so the functions are pretty much the same. Especially near infinity. Because for first to be one, a=infinity or the functions are very quickly rising(like exponential). So, the second one is also 1, which is 1. :)
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CadRipper

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« Reply #9 on: November 26, 2003, 09:00:45 am »
Quote
Originally posted by TheGeneral
He is right you know, the Hospital\'s guy rule works for any form of 1/0 or 0/1, but not infinity/infinity, I dun think. Don\'t quote me on this.

Hum, I quote you nevertheless  ;)

What I meant is that it applies to both infinity/infinity and 0/0 (it\'s only a view of the mind, both are the same) for functions that are continuous around a . That\'s precisely the point of this theorem, otherwise it wouldn\'t be possible to compute the limit so easily.

If I remember well, because it\'s more than 10 years away :D

(btw 1/0 tends to infinity and 0/1 to 0, those are trivial cases)

Quote
Originally posted by lynx_lupo
I\'ve never heard of that rule, but the way I see it, the first limit(?) is 1 and the second limit\'s parameters are just diferentials, so the functions are pretty much the same. Especially near infinity. Because for first to be one, a=infinity or the functions are very quickly rising(like exponential). So, the second one is also 1, which is 1.

No, you can\'t say the limit of infinity/infinity is 1.  For instance (bad example but I lack the time atm), 2/x and 4/x are both tending to infinity when x tends to 0. I don\'t get your point for the rest (?).

To see how the H?pital\'s theorem can be useful, take f(x)=1/x and g(x)=ln(x^2), you\'ll see it\'s worth it.

lynx_lupo

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« Reply #10 on: November 26, 2003, 01:25:57 pm »
Why not 1? I don\'t see what you\'ve tried to show with the examples. ?(
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Caldazar

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« Reply #11 on: November 26, 2003, 03:23:14 pm »
Quote
Originally posted by seperot
Quote
No way! The answer is 42!


of course the same as the answer to the meaning of life :P

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CadRipper

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« Reply #12 on: November 26, 2003, 07:13:18 pm »
Quote
Originally posted by lynx_lupo
Why not 1? I don\'t see what you\'ve tried to show with the examples. ?(

Tell me if I misunderstood what you meant.

inf / inf  is not equal to 1, but is undefined. This is tricky, but for example, let\'s consider a parameter a :

a . inf = inf, for every a  (1)

since inf is absobing regarding to the multiplication, as zero is (sorry, I don\'t know the exact words in english  ;) ). So intuitively, you get any value for inf / inf from equation (1).

That\'s what I was trying to show in my poor example : 2/x and 4/x are two functions tending to infinity when x tend to zero. But lim(f/g)=1/2, not one. You can replace 2 and 4 by any number you like.

Hey, wasn\'t this l\'H?pital a clever boy ?  :D

lynx_lupo

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« Reply #13 on: November 26, 2003, 07:32:53 pm »
Quote
Originally posted by CadRipper
Tell me if I misunderstood what you meant.

inf / inf  is not equal to 1, but is undefined. This is tricky, but for example, let\'s consider a parameter a :

a . inf = inf, for every a  (1)

since inf is absobing regarding to the multiplication, as zero is (sorry, I don\'t know the exact words in english  ;) ). So intuitively, you get any value for inf / inf from equation (1).

Well, yeah. But I don\'t agree with the last sentence. As long as there are bijections(sorry, I don\'t know the exact words in english  ;) ;) ) all infs are equal, but I don\'t think you can just \"divide\" them out from (1) and (1)(b).

Quote

That\'s what I was trying to show in my poor example : 2/x and 4/x are two functions tending to infinity when x tend to zero. But lim(f/g)=1/2, not one. You can replace 2 and 4 by any number you like.

Hey, wasn\'t this l\'H?pital a clever boy ?  :D

Hm... I see a proof of \"hospital\"\'s wrong here, then. Hm, gotta study some more, I think we\'re in what you call algebra now(groups, matrix...).

I\'ll get you back! ;)
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« Reply #14 on: November 26, 2003, 11:22:38 pm »
Well I might as well show my math geekiness here....
Yes, this rule works only for the cases of inf/inf and 0/0 .

To prove CadRipper\'s point that inf/inf does not have to equal one heres a basic example...

lim          (X^2/e^x)    is inf/inf
(x->inf)

(if that format doesn\'t make any sense... its the limit as x approaches infinity of: x squared divided by e to the power of x)

Using L\'Hopital\'s rule it becomes:

lim          (2X/e^x)    is still inf/inf
(x->inf)

And once more:

lim          (2/e^x)    is 2/inf or 0.
(x->inf)

Hopefully that clarifies things a bit..... isnt math fun? ;)
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