September 5th, 2017, 06:37 PM  #21 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,297 Thanks: 94  It's not "suggested." It's the result of the proof. See previous line.

September 5th, 2017, 07:23 PM  #22 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,235 Thanks: 2412 Math Focus: Mainly analysis and algebra 
No it's not. The proof is to show that there exists a suitable value of $\delta$ for every $\epsilon$. It's an existence proof. In fact, your work explicitly shows that there is not a single value for $\delta$ because you search for an upper bound. Moreover, your previous line actually says that the condition is $xL \lt \frac{\epsilon}{K}$. This is true if $\delta = \frac{a\epsilon}{K}$ for every $a \in (0, 1)$. So it is completely incorrect to say that the previous line in any way implies that $\delta = \frac{\epsilon}$. The next paragraph of your proof contains a similar problem, exacerbated by the use of the symbol $I$ to be both a number and an interval. Neither do you actually address the problem of what to do if you can't find an upper bound of $g(x)$ in your interval. I would say that at no stage do you actually prove that the expression you have selected to determine $\delta$ satisfies the definition of the limit. Perhaps that's a bit harsh, but it's certainly nothing like as clear a working through the definition. It's an illustration of the method we use to determine a suitable expression. But you never start with an expression for $\delta$ with you then demonstrate leads to the conclusion that $f(x)L \lt \epsilon$. You also have lines in there such as "suppose you can find $f(x)L \le xLg(x)$". Where do you show that I always can find such a factorisation? And what happens if I can't? What about when $f(x)= \frac{\sin x}{x}$? Then with $a=0$ and $L=0$ we have $f(x)L = \frac{\sin x}{x} \le x\frac1{x^2}$. In this case I've demonstrated that I can find the factorisation that you seek, but the following step cannot be completed. You also have $xL$ when you should have $xa$. Last edited by v8archie; September 5th, 2017 at 07:28 PM. 
September 5th, 2017, 10:02 PM  #23 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,297 Thanks: 94 
The Principle (Corrected from post # 19. $\displaystyle xL$ should be $\displaystyle xa)$ $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if $\displaystyle f(x)L<\epsilon$ if $\displaystyle xa<\delta$ If you can write $\displaystyle f(x)L\leqxaK$, then $\displaystyle f(x)L\leqxaK<\epsilon$ if $\displaystyle xa<\frac{\epsilon}{K} $ $\displaystyle \delta =\frac{\epsilon}{K}$ If you can write $\displaystyle f(x)L\leqxag(x)$, then 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 $\displaystyle I$, $\displaystyle g(x)\leq M$ $\displaystyle xag(x)\leqxaM$ If also $\displaystyle xa < \frac{\epsilon}{M}, xaM<\epsilon$, then $\displaystyle f(x)L\leqxag(x)\leq xaM < \epsilon$. $\displaystyle \delta = \min(I,\frac{\epsilon}{M})$ 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 7th, 2017, 09:04 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,235 Thanks: 2412 Math Focus: Mainly analysis and algebra 
I think that by now you have demonstrated that the approach I took earlier is at least as easy and understandable as what you are trying to do.


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