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I am not sure how this questions sounds but might it be possible to actually compute and not just prove the limit of any arbitrary function whose limit exists using $\epsilon-\delta$ notation?

$$\forall \epsilon >0, \exists \delta>0 :\forall x, |x-c|<\delta\implies|f(x)-L|<\epsilon $$

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    Hard to tell if something "*might be possible*," but I've never seen anything that would match this.2012-12-30
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    I think you might have to work on your logic there, it seems like only constant functions $f(x) = L$ would statify it. I suggest something like $$\forall \epsilon >0, x, \exists \delta>0, L :|x-c|<\delta\implies|f(x)-L|<\epsilon$$2012-12-30
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    I don't get what @Arthur means by $\forall \epsilon >0,x $ or actually most of it...2012-12-30
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    Yeah, I'm confused by @Arthur's point, too. Why does $L$ depend on $\epsilon$ here?2012-12-30
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    @ThomasAndrews What I mean is that as it stands, for any $\epsilon$, $f$ is closer than $\epsilon$ from $L$, for all $x$. You should rather say that given an $\epsilon$ and an $x$, there exists a $\delta$ and an $L$ (with $\delta$ depending on both $\epsilon$ and $x$, and $L$ depending on just $x$). That is what I'm trying to convey. I'm not used to logic notation conventions, so I might've skipped a qualifier or two.2012-12-30
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    $\exists \delta>0,L$ means "there exists a delta greater than zero, and an L." But $L$ is a constant which is not supposed to depend on $\epsilon$2012-12-30
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    @ThomasAndrews But it _is_ supposed to depend on $x$. Really, my problem with OP's logical statement is that $x$ seems to be qualified at an awkward place, and $L$ is not qualified at all.2012-12-30
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    No, @Arthur, it depends on $c$. We are talking $\lim_{x\to c} f(x)$.2012-12-30
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    @ThomasAndrews Oh, yes. It is the $c$ that messes with my head.2012-12-30

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The nature of the $\epsilon-\delta$ definition is that it takes in a function $f(x)$ a limit point $a$, and a limit value $L$, and provides a definition for $\lim_{x\to a} f(x) = L$. Thus the nature of the beast is not one of calculation - it requires even a proof to show that if $\lim_{x\to a} f(x)=L_1$ and $\lim_{x\to a} f(x)=L_2$ then $L_1=L_2$, because the definition does not assert that only one limit can exist.

It is best to realize that the $\epsilon$ definition of limits was arrived at as a formalism of something that all mathematicians understood. They had been using limits and continuity for centuries before they came up with this definition. The definition just finally gave a strong form to validate a limit value.