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I have to find a positive number $\delta<2$ such that $|x−2| < \delta \implies |x^2−4| < 1$.

I know that $ \delta =\frac{1}{|x+2|} $ has this behaviour, but it is not guaranteed for it to be less than two. I don't know what else to do. Thank you!

Saúl Pilatowsky-Cameo
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2 Answers2

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The choice $\frac{1}{|x+2|}$ is inappropriate for a more fundamental reason. We want a $\delta$ such that for any $x$ such that $|x-2|\lt \delta$, we have $|x^2-4|\lt 1$. So in particular $\delta$ must not depend of $x$. We now poceed to find an appropriate $\delta$.

Note that $|x^2-4|=|x-2||x+2|$. We can make $|x-2|$ small by choosing $\delta$ small, but the $|x+2|$ term could spoil things.

Suppose however that we make $\delta=\frac{1}{5}$. If $|x-2|\lt \delta$, then $2-\frac{1}{5}\lt x\lt 2+\frac{1}{5}$. Thus $x+2$ is positive and less than $5$. It follows that $|x+2|\lt 5$, and therefore if $|x-2|\lt \delta$ we have $$|x-2||x+2|\lt \frac{1}{5}\cdot 5=1.$$

André Nicolas
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  • Ok, thank you, but how did you know to select δ=1/5? – Saúl Pilatowsky-Cameo Sep 08 '14 at 03:29
  • Fooled around, asked myself how bad $|x+2|$ could be if $x$ is "near" $2$. Well, $x+2$ could well be above $4$, but it is easy to make sure it is definitely below $5$. Note that we are not asked to find the cheapest $\delta$ that will do the job. – André Nicolas Sep 08 '14 at 03:35
  • Ok, I see, thank you so much. – Saúl Pilatowsky-Cameo Sep 08 '14 at 23:22
  • @Saulpila: This was a response to someone else, who did not like the informal language I used in the comment, but has since deleted the comments. – André Nicolas Sep 08 '14 at 23:51
  • Oh, well, thank you anyway. – Saúl Pilatowsky-Cameo Sep 09 '14 at 03:41
  • You are welcome. The theme of the solution will come up again in $\epsilon$-$\delta$ and $\epsilon$-$N$ arguments. We have control of something, here $|x-2|$, can make it small. We want to show that some related quantity can be made small, often $|x-a|g(x)$. To do this, we need to make sure that by taking $|x-a|$ positive but small enough, $|g(x)|$ is bounded above. – André Nicolas Sep 09 '14 at 03:51
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Observe that : $|x^2 - 4| = |x-2||x+2| < \delta|x+2| < \delta\left(|x - 2| + 4\right) < \delta(\delta + 4) = \delta^2 + 4\delta$. Thus all we need is to solve for $\delta$: $\delta^2 + 4\delta < 1 \iff (\delta + 2)^2 < 5 \iff \delta + 2 < \sqrt{5} \iff \delta < \sqrt{5} - 2$. For example we can take: $\delta = \dfrac{\sqrt{5} - 2}{2}$

DeepSea
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