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)
Comments (19)
Votes (13)
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.
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 counterexample to disprove the resolution, it's already settled that the resolution is false. 

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.
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. 

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.
Many of the debate challenges on this site have ambiguous resolutions, or no resolution at all. At least yours had a resolution. 
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Vote Placed by bencbartlett 4 years ago
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Reasons for voting decision: Though I don't like the debate, as it seems to be worded intentionally vague in order to trick people, Con's statement about x/x always being one is simply incorrect  anyone having any sufficient background in mathematics would recognize the error of his claim. Also, the order of operations must take precedence, though this debate would have been better phrased by discussing x/x. Was it a dirty trick? Yes. Does that change the math? No. Unfortunately, Pro gets the vote.
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Vote Placed by RoyLatham 7 years ago
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Vote Placed by MTGandP 8 years ago
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Vote Placed by TheCategorical 8 years ago
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you are saying that (x6)/(x6) 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...
"(x6) 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(x6)/(x6) that would be f(x) = x for all x =/= 6 (which you agreed)
now lets look at g(x) = x(x6)/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(06)/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.
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.
If you scroll down towards the bottom to "Discontinuities which are not asymptotes"
it explains that with dividing (it uses the example of (x6).../(x6)...) 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.
o (where the "o" represents the space at 0 where x/x is undefined).
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.