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I don't understand the epsilon delta definition of a limit

"According to the formal definition above, a limit statement is correct if and only if confining x to d units of c will inevitably confine f(x) to epsilon units of L." So, if we can confine x to infinitely small delta units of c, we can confine f(x) to infitely small epsilon units of L. Like, constricting f(x) to L ?

Is that a right way of explaining what the general idea of the epsilon-delta method is ?

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    It's kind of hard to say exactly what's right or wrong in what you're stating. The logic actually goes the other way. It's like if I claim $f$ is continuous, I will have to be able to beat you by giving $\delta$ every time you give me $\epsilon$. – Tunococ Mar 14 '14 at 05:03
  • @Tunococ Mmm.. why is it backwards? I pretty much follow what was said in the quote. If we can contain $x$ to $\delta$ units of our limit point $c$, we can confine $f(x)$ to $\epsilon$ units of $L$. – nomlylni Mar 14 '14 at 05:28
  • The quoted statement is correct, but it is what happens after you have specified $\epsilon$ and $\delta$. The definition of limit involves the ability to find $\delta$ for every given $\epsilon$. – Tunococ Mar 14 '14 at 05:36

2 Answers2

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The definition I learned is

$ lim_{x \rightarrow y}{f(x)} = c $ if and only if for every $ \epsilon > 0 $, there is a $ \delta > 0 $ such that $ 0 < |x - y| < \delta$ implies $ | f(x) - c | < \epsilon $ for all $x$

This is still a little bit technical, lets see what it means. I think it's easiest to read $ 0 < |x - y| < \delta $ as "the distance between $x$ and $y$ is between $0$ and $\delta$".

If we use this interpretation, the definition becomes: $lim_{x \rightarrow y}{f(x)} = c$ if and only if for every $ \epsilon > 0 $, there is a $\delta > 0$ such that if the distance between $x$ and $y$ is between $0$ and $\delta$, the distance between $f(x)$ and $c$ is less than $\epsilon$.

This basically means that $f(x)$ gets arbitrarily close to $c$ (and I think this expression is still used sometimes as a more informal definition) without necessarily becoming $c$ (as James S. Cook pointed out). Suppose $f(x)$ does not get arbitrarily close to $c$, i.e., there is some constant $d$ such that $f(x)$ will stay away from $c$ with at least distance $d$. Then we can show the limit definition does not hold: take $\epsilon = d$. Now there doesn't exist a $\delta$ such that $0 < |x - y| < \delta \rightarrow |f(x) - c| < \epsilon $ (because we just assumed that $f(x)$ would never get closer to $c$ than distance $d$, and remember that this happening is equivalent to $|f(x) - c| < \epsilon = d$).

What may be helpful too is a (trivial) proof using the limit definition. Usually in these proofs, you take an $epsilon$, and return a $\delta$ for which you have proved that $ 0 < |x - y| < \delta \rightarrow | f(x) - c | < \epsilon $ (when dealing with continuous functions you usually just use $|x - y| < \delta$ in your proof). I prove that $\lim_{x \rightarrow y} x = y$ (or, equivalently $f(x) = x$ and $c = y$ in the original formulation).

Take $\delta = \epsilon$. Then $ |x - y| < \delta \rightarrow | f(x) - y | = | x - y | < \delta = \epsilon $. So the condition that there exists some $\delta$ for which $ |x - y| < \delta \rightarrow | f(x) - c | < \epsilon $ holds is true, as I've just shown. More advanced proof usually use a similar logic, but the expression for $\delta$ and working out the $| f(x) - c | < \epsilon$ can become quite hard. This is why other theorems are often used (for continuous functions, $ lim_{x \rightarrow y} f(x) = f(y) $, and compositions, products, sums of continuous functions are again continuous, which can help you out very often).

Also, the definition requires that something must hold for every $\epsilon$. Sometimes, teachers explain this as a game: you can choose $\epsilon$ freely, and you can give a procedure to show that something (there exists a $\delta$ such that...) will hold, then we can say $lim_{x \rightarrow y} f(x) = c$. So, if you win this game by giving such a procedure, you basically got the recipe for a proof!

If you cannot win, it suffices to give a single $\epsilon$ such that $f(x)$ will never get closer to $c$ then distance $\epsilon$.

The wikipedia page on the ($\epsilon$, $\delta$)-definition is pretty good. Also, the picture from there may be helpful when you try to visualize $\delta$ and $\epsilon$ as distances.

delta, epsilon limit-definition visualization

Ruben
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    I like your pictures and there is much goodness in here, but, I don't think the inclusion of the limit point is optional. We don't have to consider the limit point itself in the limit. In short, I think you're missing a zero. Rather than express it here, I wrote an answer to explain in detail. – James S. Cook Mar 17 '14 at 02:18
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    You need to correct your correction. When you say "without actually becoming c", you sound like it's not permitted for the function to become c. In most cases, the function does become c, but it's not required. You should either put "with or without actually becoming c" or "without (necessarily) actually becoming c". I apologize if this seems nit-picky, but limits are definitely a nit-picky part of math, and if you leave it the way you have it, it implies that the function can't equal the limit, which could confuse newbies. – msouth Mar 17 '14 at 03:48
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    Correctionception... You're definitely right though. The correction I did was just plain wrong. I blame it on the lack of sleep this night. – Ruben Mar 17 '14 at 03:56
  • @Ruben Sir Just one thing i was curious why cant the limit definition be based on" for every delta there exits epsilon thing "? Why the mathematicians give def for "every epsilon there exists delta", was there some kind of contradiction in that def which i thought of for some special functions –  Apr 03 '21 at 12:35
  • @Ruben if you had used a instead of y in teh example, cognitive load would have been lighter :) – senseiwu Jan 18 '22 at 18:55
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Let $y \in \mathbb{R}$ and $f$ a function whose domain is defined for points near $y$. We say $\lim_{x \rightarrow y} f(x)=L \in \mathbb{R}$ iff for each $\epsilon >0$ there exists $\delta >0$ such that if $x \in \mathbb{R}$ and $0 < |x-y| < \delta$ then $|f(x)-L|<\epsilon$.

Notice the $0<|x-y|$ means we do not consider the value $x=y$ for the limiting process. Now, it is often the case that $L=f(y)$ and the limit is calculated by mere evaluation of the function $f$; this is the case of a function which is continuous at $y$. In fact, this is the definition of continuity at $y$ for $f$. We say $f$ is continuous at $x=y$ iff $\lim_{x \rightarrow y}f(x) = f(y)$. Unless $y$ is an endpoint of the domain or an isolated point then some fine print applies, but let us focus on the essential point here.

One important example (there are others, but certainly this class of examples is unavoidable in the study of calculus!) of the exclusion of $x=y$ in the limit is found in difference quotients: if the limit $$ \lim_{x \rightarrow y} \frac{f(x)-f(y)}{x-y} $$ exists then we call it $f'(y)$ the derivative of $f$ at $y$. It is never reasonable to just plug $x=y$ into the difference quotient because it necessitates division by zero. So, the $0$ in the definition "$0 < |x-y|$" is of critical importance because it is the subtle point which separates taking a limit from just doing algebra. Although, I will admit, the majority of the limits we encounter in first semester calculus are just that; algebra. But, conceptually, the exclusion of the limit point from the limit is not optional in my view. It is the point.

(of course, again, for continuous expressions there is no difference, but continuity cannot be expected in many cases)

James S. Cook
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  • I'm not sure that the exclusion of $x=y$ has anything to do with the definition of the derivative. Rather, it is a matter of precision. A limit may exist without the function even being defined at that point. It "requires too much" to talk about the value of the function at that point--we're asking the question of whether the function can be accurately said to approach a certain value, and that's all we're asking. Functions with holes due to removable singularities (e.g. $x^2-1 \over x-1$) still have limits defined at the hole without us having to say anything about the value at the hole. – msouth Mar 17 '14 at 04:08
  • @msouth I agree it's not just about derivatives. I include that as an example. I totally agree there is a multitude of other examples. The reason I bring up the derivative is that it is not a weird thing like a hole in the graph. It is the central problem of differential calculus. I'll edit my answer to make this clearer, I see your comment as I read it again... thanks! – James S. Cook Mar 17 '14 at 04:48
  • @James S. Cook Sir ,Just one thing i was curious why cant the limit definition be based on" for every delta there exits epsilon thing "? Why the mathematicians give def for "every epsilon there exists delta", was there some kibd of contradiction in that def which i thought of was there some special functions where it was not valid –  Apr 03 '21 at 12:34
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    @WayBig the limit definition for functions of a real variable is commonly based on an epsilon-delta definition. There are functions for which the limit does not exist, but most functions we use allow analysis via limits. There are other approaches, for example in nonstandard analysis the limit is replaced with algebra if I understand correctly (I'm not well-versed in nonstandard analysis) – James S. Cook Apr 03 '21 at 17:05