Attachment 25363

Somehow the methodology never makes sense to me.

I understand the idea behind it, but I never understand how one chooses the appropriate number for epsilon. Can someone break down how one would do this problem?

thanks!

Printable View

- Oct 23rd 2012, 06:45 AMkingsolomonsgraveeplison delta proof problem
Attachment 25363

Somehow the methodology never makes sense to me.

I understand the idea behind it, but I never understand how one chooses the appropriate number for epsilon. Can someone break down how one would do this problem?

thanks! - Oct 23rd 2012, 07:17 AMPlatoRe: eplison delta proof problem
The fact is you do not choose the epsilon. That is given at the start of the process.

Your task is to a delta that makes it all work out.

So you control delta, in this case the size of $\displaystyle |x-3|.$

Look at$\displaystyle \left| {\frac{{2x + 4}}{5} - 2} \right| = \left| {\frac{{2x - 6}}{5}} \right| = \frac{2}{5}\left| {x - 3} \right|$.

Ah, there is that factor you control. To get $\displaystyle \frac{2}{5}\left| {x - 3} \right|<\varepsilon$ we pick $\displaystyle \delta=\frac{5\varepsilon}{2}$ - Oct 23rd 2012, 10:20 AMhollywoodRe: eplison delta proof problem
Right - you're given $\displaystyle \epsilon$ and you need to find a $\displaystyle \delta$ (which will be a function of $\displaystyle \epsilon$) that works. So you're doing the algebra backwards. Then when you write your proof, doing the algebra forward, it works. There's an excellent paper on how to do these proofs right here on the MHF Calculus forum - it's one of the "sticky" threads at the top.

- Hollywood