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September 4th, 2017, 05:49 AM   #11
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You read what I wrote, rather than what I intended to write.
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September 4th, 2017, 05:56 AM   #12
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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 $\displaystyle L, L-I<x<L+I$, and determine the maximum value $\displaystyle M$ of $\displaystyle |g(x)|$ in the interval.
Then if x is in L, $\displaystyle |g(x)|<M$.

$\displaystyle |x-L||g(x)|<|x-L|M$

If also $\displaystyle |x-L| < \frac{\epsilon}{M}, |x-L|M<\epsilon$

$\displaystyle |f(x)-L|=|x-L||g(x)|< |x-L|M < \epsilon$.

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



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

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

$\displaystyle |x+1|\left | 1+\frac{1}{x^{3}} \right |<|x+1|\left ( 1+\frac{1}{|x^{3}|} \right )<|x+1|M <\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

$\displaystyle L-I<x<L+I\rightarrow$
$\displaystyle -\frac{3}{2}<x<-\frac{1}{2}$
$\displaystyle -\frac{27}{8}<x^{3}<-\frac{1}{8}$
$\displaystyle \frac{1}{8}<|x^{3}|<\frac{27}{8}$
$\displaystyle \frac{8}{27}<\frac{1}{|x^{3}|}<8$
$\displaystyle 1+\frac{8}{27}<1+\frac{1}{|x^{3}|}<9$

The explanation and good examples similiar to above can be found at:
https://www.math.ucdavis.edu/~kouba/...ciseLimit.html
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September 4th, 2017, 06:52 AM   #13
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Quote:
Originally Posted by zylo View Post
Then if x is in L
L is a point, not an interval.

Quote:
Originally Posted by zylo View Post
$\displaystyle \delta=\min(\frac{1}{2},9)$
This is wrong.

Quote:
Originally Posted by zylo View Post
calculation of $\displaystyle M$
Why would you want to calculate $M$ after defining its value?
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September 4th, 2017, 08:19 AM   #14
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Quote:
Originally Posted by v8archie View Post
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
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September 4th, 2017, 08:22 AM   #15
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2) No it isn't see intro or link.
The statement doesn't agree with your intro either.
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September 4th, 2017, 08:52 AM   #16
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Quote:
Originally Posted by v8archie View Post
The statement doesn't agree with your intro either.
Quote:
Originally Posted by zylo View Post
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: $\displaystyle \delta=\min(\frac{1}{2},\frac{\epsilon}{9})$
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September 4th, 2017, 02:44 PM   #17
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Originally Posted by zylo View Post
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.
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September 4th, 2017, 06:28 PM   #18
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Quote:
Originally Posted by v8archie View Post
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.
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September 5th, 2017, 09:54 AM   #19
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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 $\displaystyle a$, $\displaystyle a-I<x<a+I$, and determine an upper bound $\displaystyle M$ of $\displaystyle |g(x)|$ in the interval.
Then if $\displaystyle x$ is in $\displaystyle I$, $\displaystyle |g(x)|\leq M$
$\displaystyle |x-L||g(x)|\leq|x-L|M$
If also $\displaystyle |x-L| < \frac{\epsilon}{M}, |x-L|M<\epsilon$, then
$\displaystyle |f(x)-L|\leq|x-L||g(x)|\leq |x-L|M < \epsilon$.

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


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 5th, 2017, 10:01 AM   #20
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Originally Posted by zylo View Post
$\displaystyle \delta =\frac{\epsilon}{|K|}$
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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