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December 31st, 2015, 09:43 PM  #1 
Newbie Joined: Dec 2015 From: Iraq Posts: 8 Thanks: 0  What does the epsilondelta limit definition ACTUALLY mean ? 1 How does the epsilondelta 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 epsilondelta 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: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra 

January 1st, 2016, 09:01 AM  #3  
Newbie Joined: Dec 2015 From: Iraq Posts: 8 Thanks: 0 
Great ! Quote:
 
January 1st, 2016, 09:33 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 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 $\displaystyle \epsilon$. Last edited by skipjack; January 1st, 2016 at 10:28 AM. 
January 1st, 2016, 12:46 PM  #5  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271 
Understanding epsilondelta 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:
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: 7,649 Thanks: 2630 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. 
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 epsilondelta 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 epsilondelta 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: 7,649 Thanks: 2630 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.

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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