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I am reading the $\epsilon$-$\delta$ definition of a limit here on Wikipedia.

  1. It says that

    $f(x)$ can be made as close as desired to $L$ by making the independent variable $x$ close enough, but not equal, to the value $c$.

    So this means that $f(x)$ defines $y$ or the output of the function. So when I say $f(x)$ close as desired to $L$, I actually mean the result of the calculation that has taken place and produced a $y$ close to $L$ which sits on the $y$-axis?

  2. How close is "close enough to $c$" depends on how close one wants to make $f(x)$ to $L$.

    So $c$ is actually the $x$'s that I am putting into my $f$ function. So one is picking $c$'s that are $x$'s and entering them into the function, and he actually is picking those $c$'s (sorry, $x$'s) to make his result closer to $L$, which is the limit of an approaching value of $y$?

  3. It also of course depends on which function $f$ is, and on which number $c$ is. Therefore let the positive number $\epsilon$ be how close one wishes to make $f(x)$ to $L$;

    OK, so now one picks a letter $\epsilon$ which means error, and that letter is the value of "how much one needs to be close to $L$". So it is actually the $y$ value, or the result of the function again, that needs to be close of the limit which is the $y$-coordinate again?

  4. strictly one wants the distance to be less than $\epsilon$. Further, if the positive number $\delta$ is how close one will make $x$ to $c$,

    Er, this means $\delta=x$, or the value that will be entered into $f$?

  5. and if the distance from $x$ to $c$ is less than $\delta$ (but not zero), then the distance from $f(x)$ to $L$ will be less than $\epsilon$. Therefore $\delta$ depends on $\epsilon$. The limit statement means that no matter how small $\epsilon$ is made, $\delta$ can be made small enough.

    So essentially the $\epsilon$-$\delta$ definition of the limit is the corresponding $y$, $x$ definition of the function that we use to limit it around a value? Are my conclusions wrong?

I am sorry but it seams like an "Amazing Three Cup Shuffle Magic Trick" to me on how my teacher is trying to explain this to me. I always get lost to what letters mean $\epsilon$, $\delta$, $c$, $y$, and $x$, when the function has $x$ and $y$ only.

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    $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Jul 6 '13 at 17:57
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    $\begingroup$ Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. $\endgroup$ – Zev Chonoles Jul 6 '13 at 17:58
  • $\begingroup$ thank you Zev, I will check it out $\endgroup$ – themhz Jul 6 '13 at 17:59
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    $\begingroup$ My maths professor used to describe this as a sort of game. If she gives me a value $\epsilon$ $\endgroup$ – alexwlchan Jul 6 '13 at 18:02
  • $\begingroup$ Thanx alexwlchan, and that ϵ is actually a y? or an x? In which part of the f(x)=y does it go? $\endgroup$ – themhz Jul 6 '13 at 22:54
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It does take some effort to grasp the $\varepsilon$-$\delta$ definition. You shouldn't insist on getting it all on first reading; you have to work on it somewhat patiently. It has been said that patience is a virtue; maybe this is a good example of why that would be said.

The answer to your first numbered question is "yes".

The number $c$ is on the $x$-axis, where inputs to the function live, but one must be able to use numbers arbitrarily close to $c$ as inputs, but not necessarily $c$ itelf. In the most important cases, when $c$ itself is the input, one gets $0/0$ as the output, i.e. one gets no output. And it is precisely because $f(c)$ is undefined that one must consider $\lim\limits_{x\to c}f(x)$ instead.

It is the $y$-coordinate that is to be close to $L$, the distance being less than $\varepsilon$.

$\delta$ is not equal to $x$. Rather, one must make the distance between $x$ and $c$ less than $\delta$ (but the distance should not be $0$) in order to make the distance between $y$ and $L$ less than $\varepsilon$.

The answer to #5 is no: the function by which $\delta$ depends on $\varepsilon$ is not the function by which $y$ depends on $x$. But it does depend on which function that is.

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If you are a concrete or geometrical thinker you might find it easier to think in these terms. You are player $X$ and your opponent is player $Y$.

Player $Y$ chooses any horizontal lines they like, symmetric about $L$, but not equal to it.

You have to choose two vertical lines symmetric about $c$ - these create a rectangle, with $Y$'s lines. If $f(x)$ stays within the rectangle, you win.

If you always win, whatever $Y$ does, you have a limit. If $Y$ has a winning strategy you don't.

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  • $\begingroup$ @PeteL.Clark I knew I'd read it somewhere before! I was thinking of how to explain this to my 15-year-old daughter. $\endgroup$ – Mark Bennet Jul 6 '13 at 20:37
  • $\begingroup$ I have never heard before the terms, "concrete or geometrical thinker" Thank you, I am not sure what kind of thinker I am, but your explanation does help $\endgroup$ – themhz Jul 6 '13 at 23:02
  • $\begingroup$ @Mark: I certainly didn't mean to imply that you got it from me! Probably I am not the first, nor the hundred and first, to explain $\epsilon \delta$ limits in this way. I do like it though. I am tempted to leave an answer explaining why... $\endgroup$ – Pete L. Clark Jul 7 '13 at 0:29

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