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I just had a few questions concerning the epsilon delta proof of limits. To be more precise, I always get lost at the part where we take $\delta$ to be the minimum of two real numbers. Let me provide a concrete example to work with.

Consider $\displaystyle\lim_{x \to 2}x^2 = 4$

Given $\epsilon > 0$, let $\delta = \min\{1,\displaystyle\frac{\epsilon}{5}\}$, and assume $0 < |x-2|< \delta$

Then, $|x^2 - 4| = |x+2||x-2| < 5|x-2| < 5 \cdot \displaystyle\frac{\epsilon}{5} = \epsilon$

I know we are taking $|x-2| < 1$ AND $|x-2| < \displaystyle\frac{\epsilon}{5}$ however I do not understand the following:

  1. What if it turned out that the minimum is $1$. I do not quite understand why $5|x-2| < 5 \cdot \displaystyle\frac{\epsilon}{5} = \epsilon$ would hold still.

  2. Why does it matter that we take the minimum? If one is bigger then the other wouldn't it still incorporate both conditions: $|x-2| < 1$ AND $|x-2| < \displaystyle\frac{\epsilon}{5}$? I am just imagining two open intervals where one interval is inside the other so choosing the bigger one would count towards both of them.

I don't know maybe I am just not seeing something obvious. I apologize if my questions are easy or not the best. It has been bugging me though.

PiFarmer86
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2 Answers2

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If the minimum turned out to be $1$, then $1 < \dfrac{\epsilon}{5}$, so that $\epsilon > 5$.

Then $$5|x-2| < 5(1) = 5 < \epsilon\text{.} $$ See also $\epsilon$-$\delta$ proof that $\lim\limits_{x \to 1} \frac{1}{x} = 1$. for a similar problem, and Why do we need min to choose $\delta$? for an explanation of the minimum.

Community
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Clarinetist
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A typical epsilon-delta proof appeared in textbooks is arranged in an arguably unnatural way; I provide a more natural arrangement for your reference, I believe by which you can figure out why Minimum usually appears in such a proof:

If $x \in \Bbb{R}$, then $$ |x^{2}-4| = |x-2||x+2|; $$ if in addition $|x-2| < 1$ (this is to get rid of $|x+2|$, with the bound $1$ chosen for convenience), then $|x| - 2 \leq |x-2| < 1$, implying $|x+2| \leq |x|+2 < 5$, implying $|x-2||x+2| < 5|x-2|$; given any $\varepsilon > 0$, we have $5|x-2| < \varepsilon$ if in addition $|x-2| < \varepsilon/5$. Hence, for every $\varepsilon > 0$ it holds that $|x-2| < \min \{ 1, \varepsilon/5 \}$ (which just says that $|x-2| < 1$ and $|x-2| < \varepsilon/5$) implies $|x^{2}-4| < \varepsilon$; we have proved that $\lim_{x \to 2}x^{2}=4$.

A proof like the above shows clearly how a choice of $\delta$ is made; on the contrary, if a choice of $\delta$ is written in the first place then it is not "healthy" for beginners to understand the nature of such a proof.

Note that, because the map $x \mapsto x^{2}$ can be defined at $2$, no need to put the condition $0 < |x-2|$.

Yes
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  • While I follow your proof, I am still a little iffy on the idea of taking the minimum. I sort of get it. You want $|x-2| < 1$ and $|x-2| < \displaystyle\frac{\epsilon}{5}$ so I guess if it is less then the smaller of the two, it will definitely be less then the other one then. But if I had chosen the larger of the two numbers it should still be smaller then both still right? But then we are not using minimum. – PiFarmer86 Dec 10 '15 at 15:42
  • @PiFarmer86 If $\varepsilon := 1$, then $|x-2| < 1$ does not exclude the possibility that $1/5 \leq |x-2| < 1$. – Yes Dec 10 '15 at 15:50
  • I still do not follow the minimum argument because I am not sure I understand what your saying there (maybe a pic would help me) but I feel bad bugging you about it still. Thank you so much though for helping me. – PiFarmer86 Dec 10 '15 at 21:41