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http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

I've been studying these épsilon delta proofs. In the non-linear case, he gets:

$$\delta=\min\left\{5-\sqrt{25-\dfrac{\epsilon}{3}},-5+\sqrt{25+\dfrac{\epsilon}{3}}\right\}$$

Well, I know that these $\delta$ are not equal the opposite of the other, but it has shown that $x$ must be within the range covered by these two deltas. Well, I already have bounded the $x-a$ (in this case, $x-5$) in therms of $\epsilon$, so it should work that for any given $\epsilon$, i could get only the $-5+\sqrt{25+\dfrac{\epsilon}{3}}$. Why I have to get the minimum?

user108425
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  • because u want to take within $\delta$ from $(-5)$ and since all $x's$ within the range $ (5-\sqrt{25-\dfrac{\epsilon}{3}} ,-5+\sqrt{25+\dfrac{\epsilon}{3}})$ do the job, u want to make sure you take the minimal value so that so that when you take $|x-c|<\delta$ you will still be within that interval – toufik_kh.17 Dec 14 '13 at 03:12
  • @toufik_kh.17 so why can't I just select the minimum delta value and use it in the proof? – user108425 Dec 14 '13 at 03:21
  • @toufik_kh.17 at the proof part, he assumes $-\delta < x-c < \delta$ and then substitutes $(5-\sqrt{25-\dfrac{\epsilon}{3}} ,-5+\sqrt{25+\dfrac{\epsilon}{3}})$ like delta was equal to both of them, but one is not the opposite of the other! – user108425 Dec 14 '13 at 03:33
  • the proof uses :$-5+\sqrt{25-\dfrac{\epsilon}{3}}<-\delta<x<\delta$ – toufik_kh.17 Dec 14 '13 at 03:37
  • @toufik_kh.17 so why he substitutes $5-\sqrt{25-\dfrac{\epsilon}{3}}$ in the $- \delta$ part? – user108425 Dec 14 '13 at 03:40
  • check back the proof they used $-5+\sqrt{25-\dfrac{\epsilon}{3}}$ as lower bound for $-\delta$ – toufik_kh.17 Dec 14 '13 at 03:42
  • @toufik_kh.17 I've plotted the two deltas, and the $-5+\sqrt{25-\dfrac{\epsilon}{3}}$ is always the minimum of the two deltas (when $\epsilon > 0$). Why don't simply use this delta? – user108425 Dec 14 '13 at 20:27

2 Answers2

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It is because $\delta$ has to be acceptable in the worst case. Say we are proving $\lim_{x \to 0} f(x)=L$ and for (the given) $\epsilon$ we are within $\epsilon$ over the interval $\delta \in (-1,0.1)$ The definition of limit is symmetric: it says whenever $x$ is within $\delta$ of $0$, then $|f(x)-L|\lt \epsilon$ so we have to shrink the interval to make it symmetric, so our answer should be within $\delta \in (-0.1,0.1)$. This sounds restrictive, but it is not. One can prove that symmetric limits leads to the same thing as asymmetric limits and every interval includes a symmetric interval.

Ross Millikan
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He takes the minimum so that he can use the following two inequalities. $$ \delta \leq 5-\sqrt{25-\dfrac{\epsilon}{3}} \\ \delta \leq -5+\sqrt{25+\dfrac{\epsilon}{3}} $$ The first inequality above can be written as $$ -5 + \sqrt{25-\dfrac{\epsilon}{3}} \leq -\delta. $$ So the proof uses these inequalities to get $$ -5 + \sqrt{25-\dfrac{\epsilon}{3}} \leq -\delta < x - 5 < \delta \leq -5+\sqrt{25+\dfrac{\epsilon}{3}} $$ which is the inequality he intended to get (as you can see when he worked backwards). Hopefully that clears up why he needed to take $\delta$ as the minimum of the two quantities.

Pratyush Sarkar
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  • I though $5-\sqrt{25-\dfrac{\epsilon}{3}}$ should be LESS than $\delta$, now I'm confused! – user108425 Dec 14 '13 at 12:33
  • Also, he says that the delta must be the minimum of the two, bue then in the proof, he uses the two deltas. Why??? – user108425 Dec 14 '13 at 12:59
  • No, $\delta$ has to be less than $5-\sqrt{25-\dfrac{\epsilon}{3}}$ in order to have the inequality $-5 + \sqrt{25-\dfrac{\epsilon}{3}} < -\delta$. But he also needs the inequality $\delta < -5+\sqrt{25+\dfrac{\epsilon}{3}}$ in the proof. That's why he takes the minimum of the two quantities. In the proof there is only one $\delta$, but there are two $\epsilon$. The second epsilon is a way to work around the fact that $5-\sqrt{25-\dfrac{\epsilon}{3}}$ may be undefined if the $\epsilon$ is too big. – Pratyush Sarkar Dec 14 '13 at 16:00
  • I don't understand why there's only one delta. For me, $5-\sqrt{25-\dfrac{\epsilon}{3}}$ is not the opposite of $-5 + \sqrt{25+\dfrac{\epsilon}{3}}$. I completely understood the second epsilon. I also don't understand why we have to define delta as the minimum of the two candidates, since $-5+\sqrt{25-\dfrac{\epsilon}{3}}$ is always the minimum delta (i've plotted the two ones for $\epsilon > 0$ – user108425 Dec 14 '13 at 22:35
  • @user108425 There is always one $\delta$. Also $-5+\sqrt{25-\dfrac{\epsilon}{3}}$ (as you mentioned in your last sentence) is not even in the expression for $\delta$. I think you meant $5-\sqrt{25-\dfrac{\epsilon}{3}}$.Anyway, you may be right that one of the quantity is always smaller than the other. In that case you can just choose that for $\delta$. However, there is no harm in writing $\delta$ as the minimum of the two quantities. It might be even better since you don't need to bother about which quantity is smaller and in the end we only care that a $\delta$ exists. – Pratyush Sarkar Dec 15 '13 at 00:02
  • I pretty understand this, but in the 'proof part', he uses two diferent deltas. I don't understand why :( – user108425 Dec 15 '13 at 13:49
  • @user108425 I honestly don't see where two different $\delta$ were used. Can you point out which line this was done? – Pratyush Sarkar Dec 15 '13 at 17:00
  • In the proof part, he says that we must show a $\delta$ that satisfies $-\delta < x-c < \delta$. Then he substitute the $-\delta$ and $\delta$ by this: $-5+\sqrt{25-\dfrac{\epsilon_2}{3}} < x-5 < -5+\sqrt{25+\dfrac{\epsilon_2}{3}}$. He used two diferent deltas. One is $-5+\sqrt{25-\dfrac{\epsilon_2}{3}}$ and the other is $-5+\sqrt{25+\dfrac{\epsilon_2}{3}}$ when in fact we can choose only one $\delta$ and use its opposite as $-\delta$. Cleary one is not the opposite of the other. – user108425 Dec 15 '13 at 21:24
  • @user108425 NO. He chose only one $\delta$ as $\delta=\min\left{5-\sqrt{25-\frac{\epsilon}{3}},-5+\sqrt{25+\frac{\epsilon}{3}}\right}$. Therefore we can extract two inequalities from this: $\delta \leq 5-\sqrt{25-\dfrac{\epsilon}{3}}$ and $\delta \leq -5+\sqrt{25+\dfrac{\epsilon}{3}}$. The first one can be rewritten as $-5 + \sqrt{25-\dfrac{\epsilon}{3}} \leq -\delta$. I explained this clearly in my answer. Since $-\delta < x - 5 < \delta$ we can put it altogether to get $-5 + \sqrt{25-\dfrac{\epsilon}{3}} \leq -\delta < x - 5 < \delta \leq -5+\sqrt{25+\dfrac{\epsilon}{3}}$. – Pratyush Sarkar Dec 16 '13 at 00:40
  • Now I understood. Is this exactly what he did, right? Well, I understand this now. A last question, I promise: could I use the smaller delta and complete the proof, right? – user108425 Dec 16 '13 at 01:19
  • @user108425 Yes, you can just use the smaller one as the $\delta$ which is I think $-5+\sqrt{25+\dfrac{\epsilon}{3}}$. – Pratyush Sarkar Dec 16 '13 at 02:09