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!


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

  • $\begingroup$ 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

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