Wikipedia:Reference desk/Archives/Mathematics/2007 March 13

= March 13 =

limits / L'Hôpital's rule
I 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)


 * $$1^\infty = (e^0)^\infty = e^{0 \cdot \infty}$$ so it is for the same reason that $$0 \cdot \infty$$ or $$0/0$$ are indeterminate forms. For example, $$\lim_{x \to 0} (e^x)^{1/x} = e$$ is also of of the form $$1^\infty$$. --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.

$$\lim_{x \to \infty} \frac{1}{x} * x = 1$$

Note however, that $$\lim_{x \to \infty} \frac{1}{x} = 0$$ and that $$ \lim_{x \to \infty} x = \infty $$

I hope that helps. -- Ķĩřβȳ ♥  Ťįɱé  Ø  02:15, 13 March 2007 (UTC)

And $$\lim_{x \to \infty} \frac{2}{x} \cdot x = 2$$ and $$\lim_{x \to \infty} \frac{1}{x^2} \cdot x = 0$$ 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 $$f$$, $$1^{f(x)}$$ will be identically 1 as long as $$f(x)$$ 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 $$1^{\infty}$$ is an inderterminate form is that, given $$f, g$$ such that $$\lim_{x \to x_0} f(x) = 1$$, $$\lim_{x \to x_0} g(x) = \infty$$, we cannot conclude that the limit of $$f^g$$ at $$x_0$$ is 1 a priori. Phil s 03:30, 13 March 2007 (UTC)
 * However, all that being said, it is still true that $$\lim_{z \to \infty} 1^z = 1$$. So if you were to define the symbol $$x^{\infty} = \lim_{z \to \infty}x^z$$ then it would follow that under that specific definition $$1^{\infty} = 1$$.  Thus I think whether or not $$1^{\infty}$$ 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 $$1^{\infty} = 1$$ 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. --Lambiam Talk  19:44, 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 $$1^{\infty}$$ 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? Phil s 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. Phil s 22:41, 13 March 2007 (UTC)

Polynomials and rational expressions
A 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, ST47 Talk 18:46, 13 March 2007 (UTC)
 * $$\frac{4}{xy^3} - \frac{10}{x^3y} = \frac{4x^2}{x^3y^3} - \frac{10y^2}{x^3y^3} = \frac{4x^2-10y^2}{x^3y^3} = \frac{2(2x^2-5y^2)}{x^3y^3}$$ No further simplification possible indeed.


 * $$ \frac{\frac{1}{x} +\frac{1}{2}}{1-\frac{2}{x}} = \frac{1 +\frac{x}{2}}{x-2} = \frac{\frac{x}{2}-\frac{2}{2}+2}{x-2} = \frac{1}{2} + \frac{2}{x-2} $$


 * $$ \frac{2}{x} + \frac{3+\frac{6}{x}}{2+\frac{4}{x}} = \frac{2}{x} + \frac{3}{2} $$ Here again, no further simplification possible, as you cannot simplify $$ \frac{2}{x} $$.


 * --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 $$\frac{x}{2}-\frac{2}{2}$$ and $$2$$, but for are you getting from $$\frac{\frac{x}{2}-\frac{2}{2}}{x-2}$$ to 1/2? ST47 Talk 22:43, 13 March 2007 (UTC)
 * Nevermind, you did $$\frac{\frac{x-2}{2}}{x-2}$$ and canceled out the x-2. Thanks! ST47 Talk 22:44, 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?
$$\forall{x}{\in} \mathbb{N}$$, $$\frac{1}{x} + \frac{1}{x+1} = \frac{2x+1}{x(x+1)}$$ Or, for $$n = x+1$$ we have:

$$\frac{1}{x} + \frac{1}{n} = \frac{x+n}{xn}$$

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 ♥  Ťįɱé  Ø  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 ♥  Ťįɱé  Ø  05:02, 14 March 2007 (UTC)


 * I don't know if it's what you want (it's not relevant particularly to $$n=x+1$$), but it reminds me of the harmonic mean, or of reduced mass. --Tardis 15:30, 14 March 2007 (UTC)