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Limit epsilon-delta definition vs. continuity

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  1. Oct 25, 2014 #1
    Based on the following problem from http://math.uchicago.edu/~vipul/teaching-0910/151/applyingformaldefinitionoflimit.pdf:
    [tex]
    f(x) = \begin{cases}
    x^2 &, \text{ if }x\text{ is rational} \\
    x &, \text{ if } x\text{ is irrational}
    \end{cases}
    [/tex]
    is shown to have the following limit:
    [tex]
    \lim_{x\to 1}f(x) = 1
    [/tex]
    by showing ##\lim_{x\to 1}x^2 = 1## where ##x \in \mathbb{Q}## and ##\lim_{x\to 1}x = 1## where ##x \in \overline{\mathbb{Q}}## separately.

    So, I suspect that limit does not require a gapless domain because the set ℚ of rational numbers as well as the set ##\overline{\mathbb{Q}}## of irrational numbers have gaps.

    But, why the common ϵ-δ definition of a limit ##\lim_{x\to a}f(x)=L## (##a## is neither −∞ nor ∞) never emphasizes the point that ##x## does not need to come from a gapless domain? For example, my https://www.amazon.com/dp/0073532320 simply says ##x## without explaining any property about ##x##, and therefore, I was wrongfully led to believe that ##x## must come from a gapless domain.

    I see that Wikipedia's article on precise definition of limit explicitly states that ##x \in D \subseteq \mathbb{R}##. But, I think it may also wrongfully lead people to believe that limit ##\lim_{x\to a}f(x)=L## also works when the domain is so full of gaps like the set ##\mathbb{Z}## of integers and ##a## is neither −∞ nor ∞.

    Now I am wondering: what is the precise definition of limit ##\lim_{x\to a}f(x)=L## (##a## is neither −∞ nor ∞) especially when it comes to being precise about the ##x##?

    Thank you very much.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 25, 2014 #2

    lurflurf

    User Avatar
    Homework Helper

    Please define gapless domain, I do not know what that is.
    Many of the books do not state domains most of the time. If we are being careful when taking a limit we need to check that the domain of the limit and the domain of the function are compatible. So given ε we need to find δ such that d(L,f(x))<ε for all x in D such that d(a,x)<δ.
    You make a good point about the definition being a bit silly if the are no such x.

    So you have the true statement

    $$\substack{\lim\\x\rightarrow\infty\\x\in A}\mathrm{f}(x)=\substack{\lim\\x\rightarrow\infty\\x\in B}\mathrm{f}(x)=L \leftrightarrow \substack{\lim\\x\rightarrow\infty\\x\in A\cup B}\mathrm{f}(x)=L$$

    It is not as mysterious as it seems. We check some values then check some others then we have checked them all. The important detail is the two sets A and B or in your case Q and R\Q need to have a as a limit point, otherwise it is silly. Of course they do.
     
  4. Oct 26, 2014 #3
    In my own words, by "gapless" I mean a domain such as the set ##\mathbb{R}## of real numbers. ##\mathbb{R} = \mathbb{Q} \bigcup \overline{\mathbb{Q}}##. That is, for any gap between the closest pair of rational numbers, there are infinitely many irrational numbers filling in the gap making the domain "gapless".

    But, after reading Wikipedia about limit point, I think what I want to say by "gapless" is what you say as "limit point".

    May I know the difference between "limit point" and "cluster point"? Googling for "limit point" turns up "cluster point" and it seems that some say that ##a## in ##\lim_{x\to a} f(x) = L## is a "limit point" and another a "cluster point".

    Thank you very much for answering.
     
    Last edited: Oct 26, 2014
  5. Oct 26, 2014 #4

    pasmith

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    Homework Helper

    I think the concept you are looking for is that of connectedness.

    Roughly, a subset [itex]S \subset \mathbb{R}[/itex] is disconnected if and only if there exist disjoint non-empty open sets [itex]U \subset \mathbb{R}[/itex] and [itex]V \subset \mathbb{R}[/itex] such that [itex]S \subset U \cup V[/itex]. A set is connected if and only if it is not disconnected.

    Both the rationals and the irrationals are disconnected: [tex]
    \mathbb{Q} \subset (-\infty, \sqrt 2) \cup (\sqrt 2, \infty), \\
    \mathbb{R} \setminus \mathbb{Q} \subset (-\infty, 0) \cup (0, \infty).[/tex] Indeed both are totally disconnected so that their only connected subsets are singletons: between every pair of rational numbers there is an irrational number and between every pair of irrational numbers there is a rational number.

    However, sets of the form [itex]\{ x \in \mathbb{Q} : |x - a| < R\}[/itex] always consist of more than one point when [itex]R > 0[/itex], so for every [itex]\delta > 0[/itex] there is a rational [itex]x \neq 1[/itex] such that [itex]|x - 1| < \delta[/itex]. The same is true of sets of the form [itex]\{ x \in \mathbb{R} \setminus \mathbb{Q} : |x - a| < R\}[/itex]. This makes it meaningful to consider limits through rational or irrational values.

    Contrast that with [itex]\lim_{x \to 1} f(x)[/itex] where [itex]x \in \mathbb{Z}[/itex]. [itex]\mathbb{Z}[/itex] is again totally disconnected, but obviously there is no integer [itex]x \neq 1[/itex] such that [itex]|x - 1| < \frac12[/itex], which is why we can't take that limit.
     
    Last edited by a moderator: May 7, 2017
  6. Oct 26, 2014 #5
    Many many many thanks for pointing out the concept of being connected vs. being disconnected to me. Yes, that's exactly what I want to say by my word "gapless".

    Once again, thank you very much. I really appreciate your enlightenment.
     
  7. Nov 18, 2014 #6

    Stephen Tashi

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    Science Advisor

    I think the problem does not state that the definition of limit says that the existence of two special limits implies the existence of the desired limit. It expects you to prove that the existence of those two special limits implies the result.

    Because of the existence of the two special limits, given [itex] \epsilon > 0 [/itex] there exists a [itex] \delta_1 > 0[/itex] such that [itex]0< | x - 1 | < \delta_1 [/itex] and [itex] x \in \mathbb{Q}\rightarrow |(f(x) - 1 | < \epsilon [/itex]. There also exists a [itex] \delta_2> 0 [/itex] such that [itex] 0 < |x-1| < \delta_2 [/itex] and [itex] x \in \overline{\mathbb{Q}} \rightarrow |f(x) - 1| < \epsilon [/itex].

    Let [itex] \delta = min( \delta_1, \delta_2) [/itex]. Argue that [itex] 0 < |x - 1| < \delta \rightarrow |f(x) -1 | < \epsilon [/itex] (regardless of whether [itex] x \in \mathbb{Q} [/itex] or [itex] x \in \overline{\mathbb{Q}} [/itex].
     
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