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

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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 1^{1000000000000000} is infact 1! --Agester 00:52, 13 March 2007 (UTC)

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

$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x}*x=1$

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

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

And $\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{x}\cdot x=2$ and $\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x^2}\cdot x=0$ 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 $1^{\mathrm{\infty }}$ is an inderterminate form is that, given $f,g$ such that $\underset{x\to x_0}{\lim }f(x)=1$, $\underset{x\to x_0}{\lim }g(x)=\mathrm{\infty }$, we cannot conclude that the limit of $f^g$ at $x_0$ is 1 a priori. Phils 03:30, 13 March 2007 (UTC)

However, all that being said, it is still true that $\underset{z\to \mathrm{\infty }}{\lim }{1}^z=1$. So if you were to define the symbol $x^{\mathrm{\infty }}=\underset{z\to \mathrm{\infty }}{\lim }{x}^z$ then it would follow that under that specific definition $1^{\mathrm{\infty }}=1$. Thus I think whether or not $1^{\mathrm{\infty }}$ 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 $1^{\mathrm{\infty }}=1$ 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:

(lim_x→T f(x) = lim_x→T h(x) = U and lim_x→T g(x) = lim_x→T k(x) = V) implies lim_x→T f(x)⊙g(x)= lim_x→T h(x)⊙k(x).

Under this definition, 1^∞ is an indeterminate form. (Take x⊙y = x^y, 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 $1^{\mathrm{\infty }}$ 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? 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)

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, ST47Talk 18:46, 13 March 2007 (UTC)

$\frac{4}{x{y}^3}-\frac{10}{x^{3}y}=\frac{4x^2}{x^3{y}^3}-\frac{10y^2}{x^3{y}^3}=\frac{4x^2-10y^2}{x^3{y}^3}=\frac{2(2x^2-5y^2)}{x^3{y}^3}$ 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? ST47Talk 22:43, 13 March 2007 (UTC)

Nevermind, you did $\frac{\frac{x-2}{2}}{x-2}$ and canceled out the x-2. Thanks! ST47Talk 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?

$\mathrm{\forall }x\in \mathbb{N}$, $\frac{1}{x}+\frac{1}{x+1}=\frac{2x+1}{x(x+1)}$ 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♥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 $n=x+1$), but it reminds me of the harmonic mean, or of reduced mass. --Tardis 15:30, 14 March 2007 (UTC)