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# lim(x->infinity) 1/x*x is 1

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 Voting Style: Open Point System: 7 Point Started: 2/3/2009 Category: Science Updated: 8 years ago Status: Voting Period Viewed: 5,945 times Debate No: 6742
Debate Rounds (3)

 Pro 1/x*x is 1 regardless of the value of x. Therefore the limit when x goes to infinity is still one since x represents the same infinity in both instances.Report this Argument Con If x = 2, then 1/2*2 = 1/4. 1/4 is not equal to one, therefore the proposition is false. I suspect that Pro will present a false proof of the claim 1/x*x = 1 in which there is a hidden division by zero. That's the ordinary way in which such erroneous proofs are constructed. However, there are some other ways to construct false proofs, such as constructing an equation that yields an equation with a phantom root. I'll play along so we can see what Pro has for us. Since it only takes one counter-example to disprove the resolution, it's already settled that the resolution is false.Report this Argument Pro Ok .. sorry about the confusion; the question should have stated the equation as (1/x) * x = 1. As s0m31john this does not work for x = 0, so if Con allows it I would like us to continue with the lim(x->infinity) case. The reason that zero is not relevant for me is that this came up in a discussion on probability. Where x represented the amount of possible and equally likely events. My opponent in that discussion argued that if there are an infinite amount of possible events their respective possibility is 1/infinity and therefore each and every event is impossible. My argument was that since one of the events will happen with probability (1/infinity)*infinity, that is like saying a tossed coin wont land in 50% of the cases.Report this Argument Con There is no problem with zero because (x/x) = 1. There is no convergence involved at all, (x/x) is always one because the x's cancel. I have learned a lesson about taking dopey debates in the hope they might be interesting.Report this Argument Pro Well to my defence I was looking for someone to take the opposite side. I always thought it was clear cut enough.. Hence, I took the Pro argument.Report this Argument Con Many of the debate challenges on this site have ambiguous resolutions, or no resolution at all. At least yours had a resolution.Report this Argument
19 comments have been posted on this debate. Showing 1 through 10 records.
Posted by ibstudent220 7 years ago
s0m31john has a solid point about the order of operations. Multiplication CANNOT take precedence over division, since they are essentially the same operation. Division is multiplying by the inverse of the divisor.
Posted by Ore_Ele 7 years ago
Do you realize what you are saying?

you are saying that (x-6)/(x-6) canceling to 1 with a discontinuity at x = 6 is not the same principle as x/x canceling to 1 (with a discontinuity at x = 0, that you deny).

listen to what you've already said...

"(x-6) in the denominator has a a discontinuity at x = 6, for sure." And if x is in the denominator, it has a discontinuity at x = 0, for sure, that the exact same principle as what you've already said.

the error of your assumption would imply that 0/0 = 1 or anything, not undefined.

for example f(x) = x(x-6)/(x-6) that would be f(x) = x for all x =/= 6 (which you agreed)

now lets look at g(x) = x(x-6)/x that would be g(x) = x for all x =/= 0 (which you seem to deny)

it's the same process, the same principle.

and you can easily check. If you say that x/x -> 1 (with no discontinuity)

then test g(0) = 0(0-6)/0, you say that is -6, I say it is 0/0 which is undefined.

while the limit g(x) as x -> 0 = -6, g(x) =/= -6, that is one of the key properties of limits that they teach in first year calculus.

however with x^2 = 3x, you can put x = 0 into the equation, just fine 0^2 = 0*3 -> 0=0 (correct!) That is because (as I said) the discontinuity does not exist in the original equation. the idea if somebody worked on the equation before us is not mathematical.
Posted by RoyLatham 7 years ago
So you are arguing that (x/x) has a discontinuity if it in the original equation, but it does not have a discontinuity if it is "introduced as a tool." That's a new concept of mathematics I've never encountered before. How can you tell whether or not (x/x) was in the "original" equation? Maybe (x/x) was in a prior version of the equation, but somebody canceled it out under the universal assumption that (x/x) = 1. We would have to know the history of the derivation to discover if an (x/x) had once occurred naturally as opposed to "as a tool." Nah, x's always cancel out.
Posted by Ore_Ele 7 years ago
there is a discontinuity. x/x is the same as (x-0)/(x-0) and so presents a discontinuity at x = 0. However, in the case of x^2 = 3x, there is no discontinuity because (and only because) the initial equation does not have x/x in it, it is introduced as a mathematical tool, and so can be removed as one as well. Where as the initial equation for this debate is (1/x)*x, which has the discontinuity in the initial equation, it is not introduced as a tool. Since the discontinuity is there at the start of equation, the discontinuity must be recognized at the end of the simplified equation.
Posted by RoyLatham 7 years ago
No, it is not the same principle. (x-6) in the denominator has a a discontinuity at x = 6, for sure. The question is whether an equation like x^2 = 3x can be solved without introducing a discontinuity. I claim that there is no problem dividing through by x to get x = 3. You claim that once (x/x) appears then there is a discontinuity, so the x cannot be simply canceled out. that's the issue.

To make the issue simpler, consider the equation x^2 = 0. Dividing both sides by x produces x (x/x) = 0/x. I claim that the solution to x^2 = 0 is x = 0. But by your analysis x is undefined due to the (x/x) singularity at 0.
Posted by Ore_Ele 7 years ago
http://id.mind.net...

If you scroll down towards the bottom to "Discontinuities which are not asymptotes"

it explains that with dividing (it uses the example of (x-6).../(x-6)...) out that part of the function still leaves a hole in the graph at that point. The way they get around it is by applying a piecewise function. Because if they don't, that point on the graph (even though it has a real limit) is still undefined.

The same principle holds true with x/x, it equals 1 for all x except x=0, in which it is undefined.

If this link is not enough for you, I will go home and scan my old calculus text book and post the pages for you.
Posted by RoyLatham 7 years ago
So therefore the equation x^2 = 3x cannot be solved, right? Once (x/x) is formed there is a singularity that cannot be removed, according to your conception. (x/x) remains 1 in the limit as x goes to zero, and that avoids the singularity. Cite something in the math literature that says that (x/x) does not cancel. I got through a M.S. in math without ever seeing that, so I'd like to see your support in the literature.
Posted by Ore_Ele 7 years ago
x/x = 1 for all except x=0. This should have been learn in 2nd year calculus. You'll have a graph that looks like...

--------------------o----------------- (where the "o" represents the space at 0 where x/x is undefined).
Posted by RoyLatham 7 years ago
You guys are completely out of it. x/x cancels. For example x^2/x = x; it is not undefined. If (x/x) were undefined then x*(x/x) would be undefined, and no one believes that. Similarly solve the equation x^2 = 3x Do you think the answer is x = 3, or is it undefined?
Posted by nickthengineer 7 years ago
Technically one can argue that x/x when x=0 is meaningless, not 1, 0, und, or otherwise.

7/1 = ?
To figure this out, we can ask 'what times 1 gives us 7?' Answer: 7

1/1 = ?
To figure this out, we can ask 'what times 1 gives us 1?' Answer: 1

1/0 = ?
To figure this out, we can ask 'what times 0 gives us 1?' Answer: undefined (as in nothing)

0/0 = ?
To figure this out, we can ask 'what times 0 gives us 0?' Answer: anything. everything. so nothing can be based off of x/x when x=0.
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