Timeline for Does the definition of limit correspond to the intuitive notion?
Current License: CC BY-SA 4.0
12 events
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| Dec 31, 2020 at 12:39 | comment | added | Jack L. | So to be precise, if you are only assuming that $Q$ exists and $M$ also exists such that $0<Q<M$, then you can only show that $Q\to 0$ if $M\to 0$ (If that’s what you need an answer to, then that’s it). But that is not relevant to the limit definition because for the limit definition we are not saying $Q$ should exist, rather (only) $M$ should exist and a certain inequality relating $L$ should be satisfied for every $Q$ where $0<Q<M$. | |
| Dec 31, 2020 at 12:22 | comment | added | Allawonder | The question is simple. Assume $Q$ exists, and that $M$ also exists, and $0<Q<M.$ Furthermore assume that we want to show that $Q\to 0.$ How do we do this if we insist that $M$ does not go to $0$? | |
| Dec 31, 2020 at 12:05 | comment | added | Jack L. | If no (positive) $M$ can be found (which is equivalent to $M=0$), then $Q$ cannot exist (because the interval will reduce to $0<Q<0$) $$\,$$ You’d have to understand that the significance of the limit definition is not regarding $x\to x_0$ per se but regarding whether we can find any deleted neighborhood about $x_0$ whose values for $f$ are arbitrarily close to $L$. The “any deleted neighborhood” becomes $(x_0-\delta\,,x_0+\delta)\setminus\{x_0\}$ or equivalently $0<|x-x_0|<\delta$ and “arbitrarily close to $L$” becomes $|f(x)-L|<\epsilon$. | |
| Dec 31, 2020 at 10:38 | comment | added | Allawonder | If $Q$ can approach $0$ regardless of whether or not $M$ does, then what's the point of $M$? | |
| Dec 31, 2020 at 10:31 | comment | added | Jack L. | If the language says that there exists $M$ such that whenever $0<Q<M$......, then YES, we can guarantee that $Q\to 0$ even if we do not force $M$ to be arbitrarily small ($Q$ can be arbitrarily anything lying in the open interval $(0,M)$ and $Q\to 0$ is only one aspect; we can also even have $Q\to \frac{1}{2}M$ or $Q\to M$.). $$\,$$ If the language were rather to say there exists $Q$ such that whenever $0<Q<M$.....then that would’ve been a different story. | |
| Dec 31, 2020 at 9:55 | comment | added | Allawonder | Perhaps let's forget about $\epsilon$'s and $\delta$'s and whatnots for now. Suppose we have successfully bounded a quantity $Q$ such that $$0<Q<M,$$ are you saying that we can guarantee that $Q\to 0$ even if we do not force $M$ to be arbitrarily small? | |
| Dec 30, 2020 at 6:57 | comment | added | Jack L. | As long as $\delta$ exists, then $|x-x_0|$ can be arbitrarily small; it is for that reason that the language of the definition says that there exists $\delta>0$ .... You can think of it as saying $|x-x_0|$ takes arbitrarily any value between $0$ and $\delta$, hence it can take any value arbitrarily close to $0$. If no such $\delta$ existed for some $\epsilon$, that would be equivalent to saying that $|x-x_0|$ does not exist too. $$ \,.$$ Finally, as I explained with my example in the answer above, we don’t need to show that $|x-x_0|\to 0 $ is arbitrarily as $\epsilon\to 0$. | |
| Dec 30, 2020 at 2:55 | comment | added | Allawonder | OK. Then in that way of thinking of it I suppose my question is: Can we show that $|x-x_0|$ is in fact arbitrarily small as $\epsilon$ becomes arbitrarily small? This possibility is only left open by the formal definition, but yet our intuition requires it always. Can it be shown to always be true? | |
| Dec 28, 2020 at 22:51 | comment | added | Jack L. | Our intuition is not to suggest that $\delta\to 0$; rather, our intuition is that $x\to x_0$, which is captured by the expression $0<|x-x_0|<\delta$——to wit, $|x-x_0|$ can be arbitrarily small as we please as long as the “arbitrariness” is no more than $\delta$. | |
| Dec 28, 2020 at 22:37 | comment | added | Allawonder | I think my main question is the one labelled (i). Thanks for your response, but I might not have phrased the question well. The force of the definition is that for any $\epsilon>0$ no matter how small we can find some $\delta=\delta(\epsilon)$ so that the inequality in $\epsilon$ is satisfied whenever the one in $\delta$ is. Now my question is, clearly, $\epsilon$ is assumed to be approaching $0$ here. But $\delta$ is not, as our intuition suggests it to be since we want to capture the idea that $x\to x_0.$ Can this in fact be shown from the definition itself? | |
| Dec 28, 2020 at 17:09 | history | edited | Jack L. | CC BY-SA 4.0 |
added 56 characters in body
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| Dec 28, 2020 at 17:03 | history | answered | Jack L. | CC BY-SA 4.0 |