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In the proof, why do we take the absolute value of $x_n - 2$ to be the minimum of $(1,\epsilon/5)$?

DeepSea
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SirPythagoras
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  • Because then $|x_n + 2| \lt 5$ and $|x_n - 2| \lt \frac{\epsilon }{5}$ and their product will be less than $\epsilon$ – Ishfaaq Dec 12 '14 at 02:27
  • I just don't understand what it means to be less than the minimum function. I'm having a hard time visualizing it – SirPythagoras Dec 12 '14 at 02:29
  • We want $\delta$ to be smaller than both $\epsilon $ and $1$. I'll try to find something you can read. http://math.stackexchange.com/questions/676556/scratch-work-for-delta-epsilon-proof-for-lim-x-to-13-sqrtx-4-3/676976#676976, http://math.stackexchange.com/questions/687261/help-with-epsilon-delta-definition/687323#687323, http://math.stackexchange.com/questions/687414/prove-continuity-for-cubic-root-using-epsilon-delta/687479#687479 Check these out. Comment if you need more help. – Ishfaaq Dec 12 '14 at 02:31
  • Closely related. – Cameron Buie Dec 12 '14 at 02:39

2 Answers2

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The fact, that $|x_n-2|<1$ is used in the proof. We can use another constant, but then $\epsilon$ will be divided not by 5, but by the number from the inequality similar to $3<...<5$ in your proof.

Przemysław Scherwentke
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It is not strictly necessary to make this precise choice. What one needs is to be able to bound $|x_n + 2| \ |x_n - 2|$ from above by $\epsilon$. And this choice happens to work.

If you want to find such a proof yourself it can be better to initially say $|x_n -2|$ is bounded by some (small) $\delta$. So then $4- \delta < x_n+2 < 4 + \delta$, and $|x_n + 2| \ |x_n - 2| \le 4 \delta + \delta^2$.

Now pick $\delta$ in some way so that you can show (easily) that $4 \delta + \delta^2 \le \epsilon$.

quid
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  • This makes much more sense. I was confused as to why we chose the number 5. – SirPythagoras Dec 12 '14 at 02:35
  • It seems as though it was an arbitrary choice, which just happened to work. But if we'd like to be more specific, choose delta. Okay thanks! – SirPythagoras Dec 12 '14 at 02:36
  • You are welcome. Yes there are plenty of possibilities, you could also do it with other choices, say $\min {0.5, \epsilon / 4.5}$ or $\min {2, \epsilon / 6}$ and so on. – quid Dec 12 '14 at 02:43