Wikipedia:Reference desk/Archives/Mathematics/2007 March 13
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March 13
editlimits / L'Hôpital's rule
editI was wondering, how does one prove that 1∞ is an indeterminate form? Is there a proof for it? I was always under the impression that 1 to any power was one since i learnt what an exponent was. Thanks! P.S. usually when doing limits we're told that one effective way to check our work was to plug in numbers. (put in a large number to substitute for infinity) and of course the calculator says 11000000000000000 is infact 1! --Agester 00:52, 13 March 2007 (UTC)
- so it is for the same reason that or are indeterminate forms. For example, is also of of the form . --Spoon! 01:17, 13 March 2007 (UTC)
I see. How about zero times infinity? (if you don't mind me asking). I'm sure every young scholar growing up believed anything times zero was zero! (thanks for the proof i was curious and my text book didn't explain) --Agester 01:36, 13 March 2007 (UTC)
Zero times infinity is undefined.
Note however, that and that
I hope that helps. --Ķĩřβȳ♥ŤįɱéØ 02:15, 13 March 2007 (UTC)
And and are some other possible values. That's why it's indeterminate. --Spoon! 02:54, 13 March 2007 (UTC)
- The important point here is that "inderminate forms" are just a tool to check if we are justified in computing the limit product/composition of functions as the product or composition of the limits of the functions in question. 1 to any (arbitrarily large) power will always be 1. Thus for any function , will be identically 1 as long as is defined. Similarly, 0 times any function is the constant function zero, which of course has limit zero everywhere. What we mean when we say is an inderterminate form is that, given such that , , we cannot conclude that the limit of at is 1 a priori. Phils 03:30, 13 March 2007 (UTC)
- However, all that being said, it is still true that . So if you were to define the symbol then it would follow that under that specific definition . Thus I think whether or not is indeterminate depends on how you define that symbol. Dugwiki 19:17, 13 March 2007 (UTC)
- Or, I guess to put it another way, as the article indeterminate form says, "The indeterminate nature of a limit's form does not imply that the limit does not exist". It probably makes sense in most cases to define in such a way that the limit equals one. But it is possible, under some situations like the ones mentioned previously, to obtain other values. Dugwiki 19:24, 13 March 2007 (UTC)
- If ⊙ is some binary operator, and T, U and V are values from R ⋃ {−∞, ∞}, then U⊙V is called an indeterminate form if the following implication does not hold for all real functions f, g, h and k:
- (limx→T f(x) = limx→T h(x) = U and limx→T g(x) = limx→T k(x) = V) implies limx→T f(x)⊙g(x)= limx→T h(x)⊙k(x).
- Under this definition, 1∞ is an indeterminate form. (Take x⊙y = xy, T = 0, U = 1, V = ∞, f(x) = exp(x4), g(x) = x−2, h(x) = exp(x2), g(x) = x−4.) This is independent of how the symbol is defined, and even of whether it is defined. --LambiamTalk 19:44, 13 March 2007 (UTC)
- If ⊙ is some binary operator, and T, U and V are values from R ⋃ {−∞, ∞}, then U⊙V is called an indeterminate form if the following implication does not hold for all real functions f, g, h and k:
- I guess I mispoke in the first paragraph (which is why I tried to clear it up in the second paragraph.) All I was trying to say was that even though is an indeterminate form, that doesn't mean you can't evaluate the limit. What the limit is will depend on the context. Dugwiki
- That is why I said one cannot conclude a priori. What is your point? Phils 21:35, 13 March 2007 (UTC)
- I was responding to the original poster, not implying you said anything incorrect, Phils. No need to get defensive. Dugwiki 21:51, 13 March 2007 (UTC)
- Sorry. Phils 22:41, 13 March 2007 (UTC)
- That is why I said one cannot conclude a priori. What is your point? Phils 21:35, 13 March 2007 (UTC)
- I guess I mispoke in the first paragraph (which is why I tried to clear it up in the second paragraph.) All I was trying to say was that even though is an indeterminate form, that doesn't mean you can't evaluate the limit. What the limit is will depend on the context. Dugwiki
Polynomials and rational expressions
editA few questions. First, I was given 4/(XY^3)-10/(X^3Y) and told to simplify, I multiplied the first term by X^2/X^2 and the second by Y^2/Y^2 and ended up with (4X^2-10Y^2)/X^3Y^3 - can that be simplified farther, possibly be factoring the numerator?
Second, I have ((1/X)+(1/2))/(1-(2/X)), and I multiplied by X/X to get (1+(1/2)X)/(X-2). Can this be simplified?
The third is (2/X) + (3+(6/X))/((2+(4/X)) and what I did was simplify the second term to 3/2, though I wish there was something I could do to eliminate the X in the denominator of the first term?
Thanks, ST47Talk 18:46, 13 March 2007 (UTC)
- No further simplification possible indeed.
- Here again, no further simplification possible, as you cannot simplify .
- --Xedi 22:31, 13 March 2007 (UTC)
- I'm not sure what you did in the second one, the last step. I realize that you're splitting the numerator into and , but for are you getting from to 1/2? ST47Talk 22:43, 13 March 2007 (UTC)
- Nevermind, you did and canceled out the x-2. Thanks! ST47Talk 22:44, 13 March 2007 (UTC)
- I'm not sure what you did in the second one, the last step. I realize that you're splitting the numerator into and , but for are you getting from to 1/2? ST47Talk 22:43, 13 March 2007 (UTC)
- One thing about that - you have to verify that the denominators aren't 0, or you can't cancel with them. It's a special case and doesn't matter often, but it's worth pointing out. Black Carrot 06:41, 15 March 2007 (UTC)
Is there a specific name for this rule?
edit, Or, for we have:
It seems like a component of Partial fractions. Or maybe it's just one of the many methods of getting a common denominator. Thanks. --Ķĩřβȳ♥ŤįɱéØ 23:21, 13 March 2007 (UTC)
- As a math major, I'm terrible at remembering names, so I can't help you in that department. However, the derivation for this is quite simple if you have taken Algebra 2/College Algebra. Simply multiply 1/x by (x+1/x+1) and 1/(x+1) by (x/x) in order to attain a common denominator. After that, you add the components together. In order to generalize the formula, you can use n instead of "x+1" but it is still just simply getting a common denominator. —The preceding unsigned comment was added by 71.81.19.90 (talk) 03:45, 14 March 2007 (UTC).
- Oh don't worry, I've taken Algebra 2/College Algebra. See my response to the question below =). And I guess I have the problem with remembering names as well, hence the question. --Ķĩřβȳ♥ŤįɱéØ 05:02, 14 March 2007 (UTC)
- I don't know if it's what you want (it's not relevant particularly to ), but it reminds me of the harmonic mean, or of reduced mass. --Tardis 15:30, 14 March 2007 (UTC)