September 4th, 2017, 05:49 AM  #11 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,237 Thanks: 2412 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: 1,301 Thanks: 94 
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 $\displaystyle xLg(x)< \epsilon.$ Choose an interval I about $\displaystyle L, LI<x<L+I$, and determine the maximum value $\displaystyle M$ of $\displaystyle g(x)$ in the interval. Then if x is in L, $\displaystyle g(x)<M$. $\displaystyle xLg(x)<xLM$ If also $\displaystyle xL < \frac{\epsilon}{M}, xLM<\epsilon$ $\displaystyle f(x)L=xLg(x)< xLM < \epsilon$. $\displaystyle \delta = min(I,\frac{\epsilon}{M})$ $\displaystyle \lim_{x\rightarrow 1}\frac{x^{4}+x+1}{x^{3}}=1$: $\displaystyle \left  \frac{x^{4}+x+1}{x^{3}} +1 \right =\left  \frac{x^{4}+x^{3}+x+1}{x^{3}} \right =x+1\left  \frac{x^{3}+1}{x^{3}} \right =$ $\displaystyle x+1\left  1+\frac{1}{x^{3}} \right <x+1\left ( 1+\frac{1}{x^{3}} \right )<x+1M <\epsilon $ $\displaystyle M=9$, $\displaystyle \delta=\min(\frac{1}{2},9)$ $\displaystyle (x+1)$ factored out (synthetic division). calculation of $\displaystyle M$: $\displaystyle L=1$ and let $\displaystyle I =1/2$ (to prevent $\displaystyle x$ going to 0 in the interval}, then $\displaystyle LI<x<L+I\rightarrow$ $\displaystyle \frac{3}{2}<x<\frac{1}{2}$ $\displaystyle \frac{27}{8}<x^{3}<\frac{1}{8}$ $\displaystyle \frac{1}{8}<x^{3}<\frac{27}{8}$ $\displaystyle \frac{8}{27}<\frac{1}{x^{3}}<8$ $\displaystyle 1+\frac{8}{27}<1+\frac{1}{x^{3}}<9$ The explanation and good examples similiar to above can be found at: https://www.math.ucdavis.edu/~kouba/...ciseLimit.html 
September 4th, 2017, 06:52 AM  #13 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,237 Thanks: 2412 Math Focus: Mainly analysis and algebra  L is a point, not an interval. This is wrong. 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: 1,301 Thanks: 94  Quote:
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 LI < x < L+I > xL < I  
September 4th, 2017, 08:22 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,237 Thanks: 2412 Math Focus: Mainly analysis and algebra  
September 4th, 2017, 08:52 AM  #16  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,301 Thanks: 94  Quote:
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: 7,237 Thanks: 2412 Math Focus: Mainly analysis and algebra  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: 1,301 Thanks: 94  Quote:
I gave M and a clear explanation and detailed stepbystep 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: 1,301 Thanks: 94 
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 $\displaystyle xa<\delta$ If you find $\displaystyle f(x)L\leqxLK$, then $\displaystyle f(x)L\leqxLK<\epsilon$ if $\displaystyle xL<\frac{\epsilon}{K} $ $\displaystyle \delta =\frac{\epsilon}{K}$ Suppose you can find $\displaystyle f(x)L\leqxLg(x)$ Choose an interval $\displaystyle I$ about $\displaystyle a$, $\displaystyle aI<x<a+I$, and determine an upper bound $\displaystyle M$ of $\displaystyle g(x)$ in the interval. Then if $\displaystyle x$ is in $\displaystyle I$, $\displaystyle g(x)\leq M$ $\displaystyle xLg(x)\leqxLM$ If also $\displaystyle xL < \frac{\epsilon}{M}, xLM<\epsilon$, then $\displaystyle f(x)L\leqxLg(x)\leq xLM < \epsilon$. $\displaystyle \delta = \min(I,\frac{\epsilon}{M})$ Example $\displaystyle \lim_{x\rightarrow 1} \left  \frac{x^{4}+x^{3}+x+1}{x^{3}} \right =1$ $\displaystyle \left  \frac{x^{4}+x^{3}+x+1}{x^{3}} \right =x+1\left  \frac{x^{3}+1}{x^{3}} \right =$ $\displaystyle x+1\left  1+\frac{1}{x^{3}} \right <x+1\left ( 1+\frac{1}{x^{3}} \right )<x+1M <\epsilon$ $\displaystyle M=9, \delta=\min(\frac{1}{2},\frac{\epsilon}{M})$, explained below. (x+1) factored out (synthetic division). Calculation of $\displaystyle M$ = upper bound of $\displaystyle \left ( 1+\frac{1}{x^{3}} \right )$ on $\displaystyle I$ $\displaystyle a=1$ and let $\displaystyle I =1/2$ (to prevent $\displaystyle x$ going to 0 in the interval), then $\displaystyle aI<x<a+I\rightarrow \frac{3}{2}<x<\frac{1}{2}$ $\displaystyle \frac{27}{8}<x^{3}<\frac{1}{8}$ $\displaystyle \frac{1}{8}<x^{3}<\frac{27}{8}$ $\displaystyle \frac{8}{27}<\frac{1}{x^{3}}<8$ $\displaystyle 1+\frac{8}{27}<1+\frac{1}{x^{3}}<9$ Some more examples: https://www.math.ucdavis.edu/~kouba/...ciseLimit.html 
September 5th, 2017, 10:01 AM  #20 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,237 Thanks: 2412 Math Focus: Mainly analysis and algebra  

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delta, epsilon, proof 
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