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So I'm trying to understand epsilon-delta proofs a lot more, but one thing that seems to be the case is that the rationale for picking various epsilons or deltas is obfuscated within the math itself. Maybe it's supposed to be apparent, but it isn't always so to me, so I'd like to understand.

Since $\lim\limits_{x \to a}g(x) = L$, we can say $\lim\limits_{x \to a}\frac{1}{g(x)} = \frac{1}{L}$. I know from reading up on this proof that you want to get to the point where you have

\begin{equation} \frac{1}{|L|}\frac{1}{|g(x)|}|g(x)-L| < \frac{1}{|L|}\frac{2}{|L|}\frac{|L|^2}{2}\epsilon = \epsilon, \end{equation}

which ALL seem to stem from some scratch work that leads to taking

\begin{equation} |g(x)-L| < \frac{|L|}{2}. \end{equation}

The problem is that I can't find the scratch work for this anywhere! I can get up to $\frac{1}{|L|}\frac{1}{|g(x)|}|g(x)-L|$ by myself, but past that I have no clue WHY we've chosen $\frac{|L|}{2}$. Would anyone be willing to illuminate the behind-the-scenes work being done here? Any help is appreciated!

ekorel
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  • You may find this page useful, for understanding how $\delta$ is chosen in an $\varepsilon,\delta$ proof. – Jam Dec 06 '23 at 20:55
  • that definitely helps for $\delta$, but in this proof you have to pick an $\epsilon$ as well, and the process seems a bit different. for instance, in the example, you’re told to find a version of $|x-1|$, but in the proof i’m doing there’s no way to extract $|x-a|$ without doing some weird stuff that i’m pretty sure i’m not supposed to do, haha. (i could be wrong though! obviously i don’t know everything.) – ekorel Dec 06 '23 at 23:39
  • @Mittens not really? it offers an alternative, but it still isn't really providing the rationale behind the choice of delta and epsilon, which is what i'm looking for. they just appear out of thin air, haha. – ekorel Dec 07 '23 at 04:07

1 Answers1

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The basic idea is this: you want to start by choosing some number $M$ such that, when $|g(x)-L|<M$, $g(x)$ is way from $0$; so that $\frac1{g(x)}$ is not too large. And, in fact, if $|g(x)-L|<\frac{|L|}2$, you have $|g(x)|>\frac{|L|}2$. But you are not forced to take $\frac{|L|}2$; any number greater than $0$ and smaller than $|L|$ will do. Suppose, for instance, that you begin with $|g(x)-L|<\frac{|L|}4$. Then $|g(x)|>\frac{3|L|}4$. Therefore$$\left|\frac1{g(x)}-\frac1L\right|=\frac{|L-g(x)|}{|g(x)L|}<\frac4{3L^2}|g(x)-L|.$$And so, given $\varepsilon>0$, you take $\delta>0$ such that$$|x-a|<\delta\implies|g(x)-L|<\frac34L^2\varepsilon\text{ and }|g(x)-L|<\frac{|L|}4.$$And then$$|x-a|<\delta\implies\left|\frac1{g(x)}-\frac1L\right|<\frac4{3L^2}\times\frac34L^2\varepsilon=\varepsilon.$$

José Carlos Santos
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  • thank you! this illuminates it more. in this example, what does it mean to make $\frac{1}{g(x)}$ “too large”? is it because we need $\frac{1}{|g(x)|} > \frac{1}{|L|}$? – ekorel Dec 07 '23 at 04:09
  • No. Since$$\left|\frac1{g(x)}-\frac1L\right|=\left|\frac{L-g(x)}{Lg(x)}\right|=\frac1{|g(x)|}\cdot\frac{|L-g(x)|}{|L|}$$and we want to make $\left|\frac1{g(x)}-\frac1L\right|$ small, we want that $\frac1{|g(x)|}\le K$ for some $K$. In my approach, $K$ turned out to be $\frac3{4|L|}$. – José Carlos Santos Dec 07 '23 at 07:38