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Continuity of f(x)=1/log|x|

  1. May 8, 2011 #1
    Is the function f(x) = 1/log|x| discontinuous at x=0? My book says yes. It is continuous according to me. Can somebody verify?
     
  2. jcsd
  3. May 8, 2011 #2

    arildno

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    Note that all you have to show is that given some [itex]\epsilon>0[/itex], you can always find a [itex]\delta>0[/itex], so that for any x fulfilling [itex]0<x<\delta[/itex], we have:
    [tex]\frac{1}{|\log|x||}<\epsilon[/tex]
     
  4. May 8, 2011 #3

    I like Serena

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    The function f is not defined for x=0 and this is a condition, so it is not continuous in 0.

    Note however, that if you expand the definition of f, so that f(0)=0, then it is continuous in 0.
     
  5. May 8, 2011 #4

    arildno

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    My bad.
    Of course it is discontinuous at x=0, since it isn't defined there.
    I Like Serena has pointed out how to make a continuous extension of f, a feat that is possible since the limit of f at x=0 exists.
     
  6. May 8, 2011 #5

    HallsofIvy

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    For another example, the function
    [tex]f(x)= \frac{x^2- 1}{x- 1}[/tex]
    is NOT continuous at x= 0 even though for all x except 0 it is equal to x+ 1 which is.
     
  7. May 8, 2011 #6
  8. May 8, 2011 #7

    arildno

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    No it does not.
    It shows a function where the two-sided limit at x=0 exists.
    That is not sufficient to establish continuity.
     
  9. May 8, 2011 #8

    HallsofIvy

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    Any numerical grapher can show only an "approximate" graph since it can calculate values only for a finite number of points.
     
  10. May 8, 2011 #9
    The function f(x) = 1/log|x| for all x non-zero, 0 for x=0 is continuous.
     
  11. May 8, 2011 #10

    Char. Limit

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    I assume you mean x=1 for that function. It's well-defined at x=0, it's f(0)=1.
     
  12. May 8, 2011 #11
    Thanks to all!
     
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