2
$\begingroup$

I been stuck now with this seemingly simple exercise for some time.

I need to show that:

$|x^2-4| < \epsilon$ when $0 < |x-2| < \epsilon(5+\epsilon)^{-1}$

But I'm at a loss.

I know that I somehow have to recognize that $|x^2 - 4| = |x-2||x+2| < \epsilon$ and then use this with the other inequality (for the delta) to prove these inequalities hold.

I would strongly appreciate any help.

Thanks in advance!

$\endgroup$
2
$\begingroup$

Hint: If $|x-2| < { \epsilon \over 5+ \epsilon}$ then $|x+2| \le |x-2| + 4 \le { \epsilon \over 5+ \epsilon} + 4 = 5 ({ \epsilon+4\over \epsilon +5}) <5$.

$\endgroup$
  • $\begingroup$ Thank you. Yet I'm not completely sure if I got it right from here: I wrote: $|x-2| < \epsilon(\epsilon+5)^{-1}$ and $|x+2| < 5 $ Thus $|x^2-4| = |x-2||x+2| < 5 \epsilon(\epsilon+5)^{-1} = \frac{5}{5/\epsilon+1} < \epsilon$ And then the delta satisfies that $|x^2-4| < \epsilon$ right? $\endgroup$ – Fabric Aug 28 '14 at 15:56
  • 1
    $\begingroup$ I would be more inclined to write $\epsilon { 5 \over 5+ \epsilon} < \epsilon$ as it seems more clear to me. The above is correct, as long as you have established $|x+2| < 5$. $\endgroup$ – copper.hat Aug 28 '14 at 16:29
  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. $\endgroup$ – Adam Hughes Aug 28 '14 at 17:01
  • $\begingroup$ If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. $\endgroup$ – Shaun Aug 28 '14 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.