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 September 4th, 2017, 05:49 AM #11 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra You read what I wrote, rather than what I intended to write.
 September 4th, 2017, 05:56 AM #12 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 Looks like everything done before I got here. But I did all the damn work, so The Principle Suppose you can put $\displaystyle |f(x)-L|< \epsilon$ in the form $\displaystyle |x-L||g(x)|< \epsilon.$ Choose an interval I about $\displaystyle L, L-I September 4th, 2017, 06:52 AM #13 Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra Quote:  Originally Posted by zylo Then if x is in L L is a point, not an interval. Quote:  Originally Posted by zylo$\displaystyle \delta=\min(\frac{1}{2},9)$This is wrong. Quote:  Originally Posted by zylo calculation of$\displaystyle M$Why would you want to calculate$M$after defining its value? September 4th, 2017, 08:19 AM #14 Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 Quote:  Originally Posted by v8archie 1) L is a point, not an interval. 2)This is wrong. 3) Why would you want to calculate$M$after defining its value? !) You are right. Typo. L should be I 2) No it isn't see intro or link.* 3) I gave the result of the calculation, and then the calculation (explanation of where it came from) to compartmentalize the calculation. *For the calculatiom of M to be valid, x has to be in the interval L-I < x < L+I -> |x-L| < I September 4th, 2017, 08:22 AM #15 Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra Quote:  Originally Posted by zylo 2) No it isn't see intro or link. The statement doesn't agree with your intro either. September 4th, 2017, 08:52 AM #16 Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 Quote:  Originally Posted by v8archie The statement doesn't agree with your intro either. Quote:  Originally Posted by zylo intro---$\displaystyle \delta = min(I,\frac{\epsilon}{M})$----- calculation -------------$\displaystyle M=9$,$\displaystyle \delta=\min(\frac{1}{2},9)$------------- https://www.math.ucdavis.edu/~kouba/...ciseLimit.html intro is correct:$\displaystyle \delta = \min(I,\frac{\epsilon}{M})$calculation has typo. it should be:$\displaystyle \delta=\min(\frac{1}{2},\frac{\epsilon}{9})$September 4th, 2017, 02:44 PM #17 Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra Quote:  Originally Posted by zylo I gave the result of the calculation, and then the calculation (explanation of where it came from) to compartmentalize the calculation. The bulk of the work in any any$\delta-\epsilon$proof is the determination of$M$. For that reason, I think your layout is not very clear. Likewise, your general statement of$\delta$appears somewhat out of the blue. September 4th, 2017, 06:28 PM #18 Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 Quote:  Originally Posted by v8archie The bulk of the work in any any$\delta-\epsilon$proof is the determination of$M$. For that reason, I think your layout is not very clear. Likewise, your general statement of$\delta$appears somewhat out of the blue. The bulk of the work is understanding. I gave M and a clear explanation and detailed step-by-step calculation. My expression for delta is correct. The explanation and good examples similiar to above can be found at: https://www.math.ucdavis.edu/~kouba/...ciseLimit.html Why don't you look at it- I am not inventing anything, just putting it in my own words after figuring it out. Anyone displeased with or confused by my post can go to Post #9.  September 5th, 2017, 09:54 AM #19 Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 Decided to redo my post #12 which had an egregious error in the introduction, where a is mixed up with L. And correct some typos and layout suggested by v8archie. The Principle$\displaystyle \lim_{x\rightarrow a}f(x)=L$if$\displaystyle |f(x)-L|<\epsilon$if$\displaystyle |x-a|<\delta$If you find$\displaystyle |f(x)-L|\leq|x-L||K|$, then$\displaystyle |f(x)-L|\leq|x-L||K|<\epsilon$if$\displaystyle |x-L|<\frac{\epsilon}{|K|} \displaystyle \delta =\frac{\epsilon}{|K|}$Suppose you can find$\displaystyle |f(x)-L|\leq|x-L||g(x)|$Choose an interval$\displaystyle I$about$\displaystyle a$,$\displaystyle a-I
September 5th, 2017, 10:01 AM   #20
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Quote:
 Originally Posted by zylo $\displaystyle \delta =\frac{\epsilon}{|K|}$
I think it's better to say that "this suggests $\frac{\epsilon}{|K|}$ as a value for $\delta$". It's only later that we prove that our value (or expression) for $\delta$ satisfies the definition of the limit.

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