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Can one construct a function having the following properties ?

  1. Apr 1, 2013 #1
    Is there a function [itex]f(x): \mathbb{R} \to \mathbb{R}[/itex] such that [itex]\lim_{x \to 0} x f(x) = a \neq 0[/itex].
     
    Last edited by a moderator: Apr 1, 2013
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  3. Apr 1, 2013 #2

    micromass

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    Such a function exists. What did you try already??

    What can you tell about ##\lim_{x\rightarrow 0} f(x)##. Is it possible that this limit is finite?
     
  4. Apr 1, 2013 #3

    CompuChip

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    If f(x) = 1/x then x f(x) = 1, except that doesn't work in x = 0. But maybe that will get you started.
     
    Last edited: Apr 1, 2013
  5. Apr 1, 2013 #4

    CompuChip

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    Haha this thread is getting confusing.

    In the end the answer is: yes, there is a function [itex]f(x): \mathbb{R} \to \mathbb{R}[/itex] such that [itex]\lim_{x \to 0} x f(x) = a \neq 0[/itex].
     
  6. Apr 4, 2013 #5
    Yes, the trick with limits is to substitute them with functions so you get a normal equation:

    Write xf(x) + ε(x) = g(x)

    with ε(x) and g(x) such as
    lim (x->0) ε(x) = 0 and lim (x->0) g(x) = a

    so that lim (x->0) ( xf(x) + ε(x) = g(x) ) <=> lim (x->0) xf(x) = a

    Now define a ε(x) and g(x) that satisfy those conditions and you can find a function f(x) that satisfies that limit.

    If you want x = 0 to be in the domain of f(x), you should also set the conditions ε(0) = 0 and g(0) = 0.
     
  7. Apr 4, 2013 #6

    HallsofIvy

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    The function f(x)= 1/x if [itex]x\ne 0[/itex], f(0)= 0 is a perfectly good function that maps all R, one to one, onto R, such that [tex]\lim_{x\to 0} xf(x)= 1[/tex]. Of course, a function such that [tex]\lim_{x\to 0} xf(x)[/tex] is non-zero cannot be continuous at x= 0.
     
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