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Tom Mattson

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## Main Question or Discussion Point

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?