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Math Help - A function that equals its limit (?)

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    A function that equals its limit (?)

    Show me a function f(x) such that f(x)=\lim_{x\to0}f(x)

    Actually this really more of a question I have than a challenge. I do have a solution but am challenging inquisitively. If that gets on everyone's nerves just kindly ignore this and I will post it later on as a challenging inquisition rather than an inquisitive challenge.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by rainer View Post
    Show me a function f(x) such that f(x)=\lim_{x\to0}f(x)

    Actually this really more of a question I have than a challenge. I do have a solution but am challenging inquisitively. If that gets on everyone's nerves just kindly ignore this and I will post it later on as a challenging inquisition rather than an inquisitive challenge.
    The obvious case is when f(x)=c for c\in\mathbb{R}. I'm not sure what else to suggest, since \lim_{x\to 0}f(x) should be a numerical value...
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainer View Post
    Show me a function f(x) such that f(x)=\lim_{x\to0}f(x)

    Actually this really more of a question I have than a challenge. I do have a solution but am challenging inquisitively. If that gets on everyone's nerves just kindly ignore this and I will post it later on as a challenging inquisition rather than an inquisitive challenge.
    f(x)=0.

    As Chris said though, in general this is not true. In, fact the question tells you the answer. If f(x)=\lim_{x\to x_0}f(x) for all x_0\in\text{Dom }f then f(x)=f\left(x_0\right) (assuming this function is continuous) and so it's constant.
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    Quote Originally Posted by Drexel28 View Post
    f(x)=0.

    As Chris said though, in general this is not true. In, fact the question tells you the answer. If f(x)=\lim_{x\to x_0}f(x) for all x_0\in\text{Dom }f then f(x)=f\left(x_0\right) (assuming this function is continuous) and so it's constant.
    (What is "Dom f?")

    Right so this is exactly my question. f(x) has to be a numerical value and a function for f(x)=\lim_{x\to x_0}f(x) to work.

    And yet a numerical value AND a function is exactly what we get as the roots of some second degree polynomials.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainer View Post
    (What is "Dom f?")

    Right so this is exactly my question. f(x) has to be a numerical value and a function for f(x)=\lim_{x\to x_0}f(x) to work.

    And yet a numerical value AND a function is exactly what we get as the roots of some second degree polynomials.
    \text{Dom }f is the domain. I don't see what the problem is. Suppose we have thta f(x)=\lim_{x\to0}f(x). Regardless of whether or not the function is continuous, and it must be, we have that f(x)=c for some c (specifically the value of the limit) if the limit exists. I don't get the reference to quadratics
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    f(x)\equiv y

    The roots of y in the equation y^2x^2+y(2x^2-1)+x^2-1=0 (which is the equation plotted in the attachment)
    are -1 and \frac{1}{x^2}-1

    So, \lim_{x\to\infty} y i.e. \lim_{x\to \infty}\frac{1}{x^2}-1=-1

    (The limit was as x goes to infinity, not 0. REALLY SORRY about that.)

    and yet the other root of y euals this limit, equals -1. So, can't it be said that the function y equals its limit?
    Attached Thumbnails Attached Thumbnails A function that equals its limit (?)-clipboard01.jpg  
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainer View Post
    f(x)\equiv y

    The roots of y in the equation y^2x^2+y(2x^2-1)+x^2-1=0 (which is the equation plotted in the attachment)
    are -1 and \frac{1}{x^2}-1

    So, \lim_{x\to\infty} y i.e. \lim_{x\to \infty}\frac{1}{x^2}-1=-1

    (The limit was as x goes to infinity, not 0. REALLY SORRY about that.)

    and yet the other root of y euals this limit, equals -1. So, can't it be said that the function y equals its limit?
    When you say f(x)=\lim_{x\to x_0}f(x) that literally means " f(x) is a function that takes the value of \lim_{x\to x_0}f(x) at every point of its domain."
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    Quote Originally Posted by Drexel28 View Post
    When you say f(x)=\lim_{x\to x_0}f(x) that literally means " f(x) is a function that takes the value of \lim_{x\to x_0}f(x) at every point of its domain."
    Thanks. And yet this is true, if it is true that f(x)=-1 AND f(x)=\frac{1}{x^2}-1

    And yet it is untrue, as you point out, for the very same reason. In other words this is a logical contradiction. And a contradiction on two counts: 1) because one root is a constant, and the other is a function, and yet both roots must be defined at the outset as EITHER a function OR a constant, not both; and 2) because the domain of one root is undefined or mathematically meaningless vis-a-vis the domain of the other root (because it involves 1/infinity).

    How does one resolve the contradiction? How do you qualify "AND" to resolve it?
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    Any continuous function has the property that \lim_{x\to x_0}f(x)= f(x_0), that's the definition of "continuous". It is true for all x_0 for any function, f, that is continuous for all real numbers.

    That is NOT true for f(x)= \frac{1}{x^2}- 1 because that function is not defined, and so not continuous, at x= 0.

    I have no idea what you are trying to say.
    Last edited by Jhevon; January 14th 2010 at 11:40 AM. Reason: fixed LaTeX
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    Quote Originally Posted by HallsofIvy View Post
    I have no idea what you are trying to say.
    Surely, the problem is rather that I am not grasping the full import of what you are all saying. So, thanks for your patience and comments. I will ruminate on the input I've received in this thread and will no doubt eventually come around to a proper understanding (both of what you're saying and of what I'm trying to get at).
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