My Math Forum What does the epsilon-delta limit definition ACTUALLY mean ?

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 December 31st, 2015, 09:43 PM #1 Newbie   Joined: Dec 2015 From: Iraq Posts: 8 Thanks: 0 What does the epsilon-delta limit definition ACTUALLY mean ? 1- How does the epsilon-delta limit definition help us prove that a limit exists? I'm not sure how this precise definition agrees with the informal definition (you can get f(x) as close as you desire to L, provided you get x sufficiently close to a). I mean, if we can show that | f(x) - L | < epsilon IF 0 < | x - a | < delta what are we actually doing? Does this show that the function in question has a definite direction, and that no matter how far we are from L, we will always get there if x gets closer and closer? If my understanding is correct, then 2- Why do some people claim that limits work only when x is very close to a? Isn't the epsilon-delta definition of limits showing that, it doesn't matter how close (x) is to (a), we can always show that a limit exists if we can provide a delta for a given epsilon, without being restricted to only small deltas? Last edited by skipjack; January 1st, 2016 at 10:31 AM.
 December 31st, 2015, 10:03 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra The limit definition is that, given an $\displaystyle \epsilon > 0$ there exists a $\displaystyle \delta > 0$ with the property you describe. It is the fact that this $\displaystyle \delta$ always exists, not matter how small (or large) $\displaystyle \epsilon$ may be that gives the concept its power. A function in one variable has no direction, so I don't know what you are trying to say. However, the idea is that, since there exists a suitable $\displaystyle \delta$ for every $\displaystyle \epsilon$, not matter how ludicrously small the latter may be, there is always an interval for which the function is at least as close to the limit as the arbitrarily specified number $\displaystyle \epsilon$. Under this description, it is perhaps easier to see why people focus on the very small $\displaystyle \epsilon$. If $\displaystyle \epsilon$ is large the result is neither impressive nor useful. However, it is wrong to say that limits only work when $\displaystyle x$ is very close to $\displaystyle a$, although the size of $\displaystyle \delta$ is often limited in proofs to a relatively small value for practical reasons, and in practice by vertical asymptotes. However, for any function that is bounded for finite values of $\displaystyle x$ we can choose very large values for large enough $\displaystyle \epsilon$. Thanks from Moamen
January 1st, 2016, 09:01 AM   #3
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Great !

Quote:
 A function in one variable has no direction, so I don't know what you are trying to say.
What I am trying to say is that, for me, the epsilon delta definition is saying that since we can show that f(x) will eventually be close to L if x will be close to a with ANY Epsilon, then we are proving that f(x) will be heading toward L no matter where we start from if we get x closer and closer to a. Is this how the definition helps us prove that a limit exists ?

 January 1st, 2016, 09:33 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra I can see what you are thinking and it can be a useful idea, but it's not exactly what the definition states. For example, suppose that we have $\displaystyle f(x)=x$ and $\displaystyle a=0$, then for $\displaystyle \epsilon = 10$ we can pick any $\displaystyle \delta < 10$. So we could choose $\displaystyle \delta = 1$. Then for $\displaystyle \epsilon = 5$ we could pick $\displaystyle \delta = 2$ and our delta is not decreasing. But that doesn't matter because the fact that a suitable $\displaystyle \delta$ exists (for every $\displaystyle \epsilon$) is sufficient. The theory doesn't imply any motion. Of course, in this example, $\displaystyle \delta$ would eventually decrease, but that isn't true for all functions. In particular, for $\displaystyle f(x)=0$, we can select any $\displaystyle \delta$ we like for each $\displaystyle \epsilon$. We can even make $\displaystyle \delta$ increase as a function of $\displaystyle \epsilon$. Thanks from Moamen Last edited by skipjack; January 1st, 2016 at 10:28 AM.
January 1st, 2016, 12:46 PM   #5
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Understanding epsilon-delta goes back to the presentation you were given to explain it.

You can take a geometric approach or an algebraic approach

I cannot state v8Archie's statement in post#2 any more clearly, but

states this as a fact without justification

Quote:
 The limit definition is that, given an ϵ>0 there exists a δ>0 with the property you describe. It is the fact that this δ always exists, not matter how small (or large) ϵ may be that gives the concept its power.
What justification did your course offer for this statement?

Do you understand what a sequence is and in particular a null sequence?

 January 1st, 2016, 01:42 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra It's also worth pointing out that $\displaystyle \lim_{x \to a} f(x) = L$ does not mean that "we will always get there if $\displaystyle x$ gets closer and closer". A limit as $\displaystyle x \to a$ is entirely independent of what happens at $\displaystyle a$ and the function need never attain the value $\displaystyle L$ anywhere. But the definition of the limit says that you can find an interval in which $\displaystyle f(x)$ is within $\displaystyle \epsilon$ of that limit, no matter how small $\displaystyle \epsilon$ is. If the function is continuous and both $\displaystyle a$ and $\displaystyle L$ are finite, then $\displaystyle f(a)=L$ and the function is guaranteed to reach the limit. Thanks from Moamen
 January 1st, 2016, 09:26 PM #7 Newbie   Joined: Dec 2015 From: Iraq Posts: 8 Thanks: 0 Thank you v8archie ! Well, I didn't mean that delta will be decreasing, neither did I mean that f(x) would eventually be equal to L. I apologize for my ambiguous explanation. I understand what the epsilon-delta definition says, but I can not relate it to the informal definition of limits. I mean, we've managed to show that | f(x) - L | < epsilon IF 0 < | x - a | < delta , for any epsilon. So what ? what does that translate to in terms of proving a function has a limit? The only way that I see how the epsilon-delta definition proves that a limit exists, is by imagining that a function, whose limit L exists, has a specific pattern (let's not use the word direction again) and it follows that pattern to get somewhere close to L as x gets somewhere close to a. And that when we say that we can provide a delta for any given epsilon, we are simply saying that we can show that the function follows that same pattern for any range around L, doesn't matter where we look at it from (big or small epsilon), it will definitely keep going by that pattern for all values of x around a. So we know what the function be near L because we know how it behaves around L.
 January 1st, 2016, 09:31 PM #8 Newbie   Joined: Dec 2015 From: Iraq Posts: 8 Thanks: 0 Thanks studiot ! I always fail to understand the obvious things. And in this case, like I said above, I know what the official definition means ,by itself, but I can't understand how it is linked to the more intuitive informal definition. I haven't got to sequences yet.
 January 1st, 2016, 10:19 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Essentially it jut says that however close you want value of the function to get to the limit, there is a set of points that will do the job. Thanks from Moamen
 January 1st, 2016, 11:04 PM #10 Newbie   Joined: Dec 2015 From: Iraq Posts: 8 Thanks: 0 That's exactly what I was looking for !

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