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September 5th, 2017, 07:37 PM   #21
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Quote:
Originally Posted by v8archie View Post
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.
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September 5th, 2017, 08:23 PM   #22
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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.
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September 5th, 2017, 11:02 PM   #23
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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 $\displaystyle a$, $\displaystyle a-I<x<a+I$, and determine an upper bound $\displaystyle M$ of $\displaystyle |g(x)|$ in the interval.
Then if $\displaystyle x$ is in $\displaystyle I$, $\displaystyle |g(x)|\leq M$
$\displaystyle |x-a||g(x)|\leq|x-a|M$
If also $\displaystyle |x-a| < \frac{\epsilon}{M}, |x-a|M<\epsilon$, then
$\displaystyle |f(x)-L|\leq|x-a||g(x)|\leq |x-a|M < \epsilon$.

$\displaystyle \delta = \min(I,\frac{\epsilon}{M})$


Example

$\displaystyle \lim_{x\rightarrow -1} \left | \frac{x^{4}+x^{3}+x+1}{x^{3}} \right |=-1$

$\displaystyle \left | \frac{x^{4}+x^{3}+x+1}{x^{3}} \right |=|x+1|\left | \frac{x^{3}+1}{x^{3}} \right |=$
$\displaystyle |x+1|\left | 1+\frac{1}{x^{3}} \right |<|x+1|\left ( 1+\frac{1}{|x^{3}|} \right )<|x+1|M <\epsilon$
$\displaystyle M=9, \delta=\min(\frac{1}{2},\frac{\epsilon}{M})$, 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
$\displaystyle a-I<x<a+I\rightarrow
-\frac{3}{2}<x<-\frac{1}{2}$
$\displaystyle -\frac{27}{8}<x^{3}<-\frac{1}{8}$
$\displaystyle \frac{1}{8}<|x^{3}|<\frac{27}{8}$
$\displaystyle \frac{8}{27}<\frac{1}{|x^{3}|}<8$
$\displaystyle 1+\frac{8}{27}<1+\frac{1}{|x^{3}|}<9$

Some more examples:
https://www.math.ucdavis.edu/~kouba/...ciseLimit.html
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September 7th, 2017, 10:04 AM   #24
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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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