PlaneShift
Fan Area => The Hydlaa Plaza => Topic started by: Kuiper7986 on November 26, 2003, 12:47:35 am
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Is this formula true?
L\'H?pital\'s rule:
If (http://www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo09.gif) is of the form (http://www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo11.gif), and if (http:////www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo12.gif) exists, then (http://www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo13.gif).
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the answers 3
dont ask how i know but trust me
its 3
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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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Originally posted by seperot
the answers 3
dont ask how i know but trust me
its 3
No way! The answer is 42!
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No way! The answer is 42!
of course the same as the answer to the meaning of life :P
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Originally posted by Kuiper7986
Is this formula true?
L\'H?pital\'s rule:
If (http://www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo09.gif) is of the form (http://www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo11.gif), and if (http:////www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo12.gif) exists, then (http://www.hsd.k12.or.us/schools/ghs/staff/abel/images/math/theo13.gif).
True, of course, and it also applies to the form 0/0. Note the recursive possibility.
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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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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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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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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)
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.
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Why not 1? I don\'t see what you\'ve tried to show with the examples. ?(
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Originally posted by seperot
No way! The answer is 42!
of course the same as the answer to the meaning of life :P
How many roads must a man walk down?
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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
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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).
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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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? ;)