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Does the limit exist if the domain is restricted?

  1. Sep 29, 2013 #1
    In the case of a function like [itex]\sqrt{x}[/itex], where the function is only defined for values equal or greater than 0, does the limit exist at 0? I know both the left and right side limits as x approaches a must be equivalent and real for the limit at a to exist, but what if the domain is restricted? Do we consider the function differently when it is on a restricted domain? Would the limit for the above exist, and equal 0, as x approaches 0?

    Sorry if this question has been asked already. I just keep reading contradictory statements online, and that it does exist only if you consider complex numbers (I am not, in this case). Any light shed on the topic would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Sep 29, 2013 #2

    mathman

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    If the domain of interest is limited, and you are concerned about an end point, then the limit is one-sided. Don't get too hung up on the wording of definitions. What is important is to understand their meaning.

    The definition for limit as described is a narrow one, only useful for one dimensional variables and in an interior position.
     
  4. Sep 29, 2013 #3

    pasmith

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    Yes. When we write
    [tex]
    \lim_{x \to a} f(x)
    [/tex]
    we are implicitly restricting [itex]x[/itex] to lie in the domain of [itex]f[/itex].
     
  5. Sep 29, 2013 #4

    arildno

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    That really depends on how we choose to give the actual limit definition, and that varies between authors.

    For example, you could form a slightly deviant e-d-definition than the normal one, saying that L is a limit of a function at x=a, so that for any epsilon>0, we can find a d>0, so that there are NO x within 0<|x-a|<d that makes |f(x)-L|>e.

    If therefore, we have an isolated point x=a on which f is defined, in a region where f is NOT defined at all, it follows that EVERY number L is a limit of f at a. Because such x's trivially don't exist there..., and hence cannot make |f(x)-a|>e, for whatever L you pick. (I haven't made this up, mymath professor used it to show us logical subtleties)

    In this way of defining a limit, it follows that uniqueness of limit only exists if the function is defined in a neigbourhood of x=a.
     
  6. Sep 29, 2013 #5
    Nonsense, if you don't mind my being direct. If the domain is restricted, there's no such thing as a one-sided limit.

    If the domain is the unit interval [0,1], would you say that the limit of 1/n as n goes to infinity is a one-sided limit? Of course not.

    In fact it's only when the domain is not restricted that we can have 1-sided limits. If I define

    f(x) = 1/x for x >= 0, and

    f(x) = 200 if x < 0, then f has two different one-sided limits; namely 200 if x -> 0 from the left; and 0 if x -> 0 from the right.

    But if I define f as above and then say, "Let us restrict our domain to x >= 0, then there is a limit, period. The question of a one-sided limit doesn't arise, because the part of the domain you are ignoring doesn't exist.

    This is analogous to fact that the definition of a closed set is always relative to a given topology. In the real numbers, the open interval (0,1) is an open set and not a closed set.

    In the topological space (0,1), the interval (0,1) is both open and closed. That's because if our universe is given as (0,1), the larger ambient space that we secretly know about in the back of our minds, actually does not exist in the context of the question. The topological space (0,1) is the entire space. It confuses things to think "Oh it's really living in the reals," because then it's harder to see that (0,1) is a closed set.

    Same deal with limits. Restricting your space makes one-sided limits into limits, because you're throwing out the "bad part" of the space.
     
    Last edited: Sep 29, 2013
  7. Sep 29, 2013 #6

    Office_Shredder

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    Is your claim that
    [tex] \lim_{x\to 0^+} \sqrt{x} [/tex]
    does not exist on the basis that I should have written
    [tex] \lim_{x\to 0} \sqrt{x}[/tex]?

    I think you'll find yourself very alone in claiming that the first limit is wrong in any way.

    Your example is very confusing because you say your domain is [0,1] but then you take a limit as n goes to infinity -you seem to have confused domain and codomain I think.
     
  8. Sep 29, 2013 #7
    Both exist, of course.

    I believe Mathman's statement was exactly the opposite of what is true. Or at the very least, confusing and;or misleading. Alone or not, that's my opinion.

    Good point. I can fix that with my function f as above, defined on [-1, +infinity). You have different one-sided limits and no limit at zero. But f restricted to [0, +infinity) does have a limit, namely 0. It's expanding the domain that introduces the possibility of two one-sided limits that differ, not restricting it. That's how I understood MM's statement.
     
  9. Sep 30, 2013 #8

    HallsofIvy

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    A perhaps more important example is [tex]\lim_{x\to 3}\frac{x^2- 9}{x- 3}= 6[/tex] where the limit exist even though the function [tex]f(x)= \frac{x^2- 9}{x- 3}[/tex] is not defined at x= 3.
     
  10. Sep 30, 2013 #9
    A function has its domain as part of the definition. Suppose I write [itex]f:X\to \mathbb R[/itex] for some [itex]X\subseteq \mathbb R[/itex]. Then the domain [itex]X[/itex] is part of the information. If [itex]a[/itex] belongs to the closure of [itex]X[/itex] and [itex]c\in \mathbb R[/itex] (I'll stay away from infinite limits, because that's not the main issue here), then there's a completely unambiguous meaning for the statement, [tex]\lim_{x\to a} f(x)=c.[/tex] It means that for every [itex]\epsilon>0[/itex], there exists a [itex]\delta>0[/itex] such that for any [itex]x[/itex] from [itex]X[/itex] with [itex]|x-a|<\delta[/itex], we have [itex]|f(x)-c|<\epsilon[/itex].

    So the definition of limit is unambiguous, as long as the domain of the function is clear. If I haven't told you what the function's domain is, then I haven't actually told you what the function is.

    As for one-sided limits... Fix [itex]f:X\to\mathbb R[/itex] and [itex]a[/itex] in the closure of [itex]X[/itex]. Then let [itex]X^+=\{x\in X: x>a\}[/itex] and let [itex]f^+:X^+\to \mathbb R[/itex] be the restriction of [itex]f[/itex] to [itex]X^+[/itex]. Then we write [itex]\lim_{x\to a^+} f(x)=c[/itex] as another way of saying [tex]\lim_{x\to a} f^+(x)=c.[/tex] Of course, this is a weaker requirement than [itex]\lim_{x\to a} f(x)=c[/itex].
     
  11. Sep 30, 2013 #10

    Office_Shredder

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    This isn't an example, because the limit definition never has the option of f(a) being evaluated. If it did, then any function which wasn't continuous at a wouldn't have the limit at a exist
     
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