Author Topic: Epsilon-Delta limit method  (Read 522 times)

0 Members and 1 Guest are viewing this topic.

Offline MathWizardTopic starter

  • Super Contributor
  • ***
  • Posts: 1622
  • Country: ca
Epsilon-Delta limit method
« on: January 19, 2023, 11:30:47 am »
I'm reviewing some early calculus again, like limits, for doing more stuff with step and impulse functions. And I'm wondering if there's anything wrong with doing it this way, since I don't remember people doing it this way, and neither my calculus books or pdfs do it directly. And I lost my real analysis book years ago.

So for finite limits anyways, since the definition uses the assumed limit L, and we know x-->c , so we know c.
So I'm been just solving some directly plugging x=c+delta, into the |f(x)-L| < e, and it works so far.

https://www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf
So in problem 4 on the bottom of page 3, you would get the polynomial  | (delta)^2 +5delta | < e
Which solves for any chosen positive e, you'll get 1 good delta answer.

And for positive delta and epsilon, the pdf works out the same thing basically, and says

|x-2|=delta < (e/6)  ==> 6delta < e (or equal)

so with the substitution way , so long as delta is between [0,1], I have

(delta)^2 +5delta < 6delta (or equal) and that's always true. And it worked for the negative delta aswell.

And thats way shorter that counting/ordering number line way


For something like 1/x, I'd say treat L as finite , L=1/delta and try induction type steps and say it goes to infinity. But I was only going to review improper integrals.

« Last Edit: January 19, 2023, 11:38:24 am by MathWizard »
 

Offline dmowziz

  • Contributor
  • Posts: 40
  • Country: ca
Re: Epsilon-Delta limit method
« Reply #1 on: January 21, 2023, 10:20:48 pm »
Hi MathWizard,

Firstly, I think maybe there is a mistake in the attached pdf...

The 1st attached image (between page 3 and 4)...
It says that |x + 3| < 6... If this is the case and epsilon (e) is positive then the last equation in my attached image is not correct.. Yeah?
That is |x + 3| < 6  does not mean that (1/ |x+3|) < 1/6.... It is greater. Yes?



Am I missing something?
I most likely am

« Last Edit: January 21, 2023, 10:54:08 pm by dmowziz »
 

Offline ledtester

  • Super Contributor
  • ***
  • Posts: 3248
  • Country: us
Re: Epsilon-Delta limit method
« Reply #2 on: January 22, 2023, 01:45:18 am »
Hi MathWizard,

Firstly, I think maybe there is a mistake in the attached pdf...

The 1st attached image (between page 3 and 4)...
It says that |x + 3| < 6... If this is the case and epsilon (e) is positive then the last equation in my attached image is not correct.. Yeah?
That is |x + 3| < 6  does not mean that (1/ |x+3|) < 1/6.... It is greater. Yes?



Am I missing something?
I most likely am



The solution just goofed writing out the inequality -- the \$\frac{\epsilon}{|x+3|}\$ and \$\frac{\epsilon}{6}\$ terms should be swapped. They just got done saying that if \$|x-2| < \frac{\epsilon}{6}\$ then \$|x-2| < \frac{\epsilon}{|x+3|}\$ because \$6 > |x+3|\$.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf