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September 5th, 2017, 07:37 PM   #21
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 Originally Posted by v8archie 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.
It's not "suggested." It's the result of the proof. See previous line.

 September 5th, 2017, 08:23 PM #22 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,116 Thanks: 2369 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 $|x-L| \lt \frac{\epsilon}{|K|}$. 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 $|f(x)-L| \le |x-L||g(x)|$". 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 $|x-L|$ when you should have $|x-a|$. Last edited by v8archie; September 5th, 2017 at 08:28 PM.
 September 5th, 2017, 11:02 PM #23 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,214 Thanks: 91 The Principle (Corrected from post # 19. $\displaystyle |x-L|$ should be $\displaystyle |x-a|)$ $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if $\displaystyle |f(x)-L|<\epsilon$ if $\displaystyle |x-a|<\delta$ If you can write $\displaystyle |f(x)-L|\leq|x-a||K|$, then $\displaystyle |f(x)-L|\leq|x-a||K|<\epsilon$ if $\displaystyle |x-a|<\frac{\epsilon}{|K|}$ $\displaystyle \delta =\frac{\epsilon}{|K|}$ If you can write $\displaystyle |f(x)-L|\leq|x-a||g(x)|$, then Choose an interval $\displaystyle I$ about $\displaystyle a$, \$\displaystyle a-I
 September 7th, 2017, 10:04 AM #24 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,116 Thanks: 2369 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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