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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_1-L_2[/tex]
[tex]L_1=L_2[/itex]
Therefore, the limit is unique.
Can you spot the flaw in the argument?
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_1-L_2[/tex]
[tex]L_1=L_2[/itex]
Therefore, the limit is unique.
Can you spot the flaw in the argument?