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Can you spot the fallacy? 
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#1
Jun1708, 03:19 PM

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I gave an extra credit problem to my Calculus I course. I told them I would give them 10 bonus points if they could prove that for a function [itex]f(x)[/itex], its limit as [itex]x \rightarrow c[/itex], if it exists, is unique. I gave them a couple of hints and told them that they would definitely have to use the definition of a limit. One student came up with the following argument which I hadn't anticipated. It's astonishingly simple and (I thought) quite clever, but it is not a proof. Here goes.
Assume the following: [tex]\lim_{\substack{x\rightarrow c}} f(x)=L_1[/tex] [tex]\lim_{\substack{x\rightarrow c}} f(x)=L_2[/tex] Now consider the following: [tex]\lim_{\substack{x\rightarrow c}} f(x)+\lim_{\substack{x\rightarrow c}} f(x)=L_1 + L_1[/tex] [tex]\lim_{\substack{x\rightarrow c}} f(x)+\lim_{\substack{x\rightarrow c}} f(x)=L_1 + L_2[/tex] Now subtract the second equation from the first to obtain: [tex]0=L_1L_2[/tex] [tex]L_1=L_2[/itex] Therefore, the limit is unique. Can you spot the flaw in the argument? 


#2
Jun1708, 03:26 PM

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Apart from that fact that it assumes that limits are unique?



#3
Jun1708, 03:28 PM

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in the beginning, by saying it equals L1 and then it equals L2 you are assuming that they are not equal. but then when you subtract them to get 0=L1L2 you are now assuming that they are equal so that when you subtract them you get 0... you can not change your assumptions in a proof.



#4
Jun1708, 03:29 PM

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Can you spot the fallacy?
I prefer to not stick my head above multiplication table, but I think left side zeroes only if L1 = L2, which has yet to be proven.



#5
Jun1708, 03:37 PM

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Can you say that the limit as x approaches c of f(x) is unique because by definition, in a function, for every x there is a unique y?



#6
Jun1708, 03:58 PM

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Of course, you guys spotted it right away. I asked her, "How do you know you get zero on the left hand side?" And she said, "Because when I subtract those limits are all equal to each oth" Then she got it, too. 


#7
Jun1708, 04:11 PM

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Also, if we start with this:
[tex]L_1 = \lim_{\substack{x\rightarrow c}} f(x) = L_2[/tex] 


#8
Jun1708, 04:16 PM

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#10
Jun1708, 04:32 PM

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My previous answer was : "Can you say that the limit as x approaches c of f(x) is unique because by definition, in a function, for every x there is a unique y?" What if I added: If there is a vertical asymptote, then the limit will either be either positive or negative infinite, so it's unique, or it will not exist because one side will tend to positive infinite while the other will tend to negative infinite and in that case, the limit will not exist but that's okay because one of the conditions you gave is that the limit exists. And if there is a hole, then the limit would be unique because it would be f(c) assuming that the function was defined at c and if the limit didn't exist, it would be because the function wasn't continuous, which is again okay because one of the conditions you gave is that the limit exists. 


#11
Jun2208, 08:55 PM

P: 122

From your original post what impressed me was not that the argument was "clever" but rather that the student seemed talented. Again, what a joy! A talented student! Although, the only student I ever had in my math classes who was not a joy just also happened to be incrediably talented. Deacon John 


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