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How to show that this function is continuous at 0?

  1. Jul 4, 2012 #1
    1. The problem statement, all variables and given/known data

    For all real numbers, f is a function satisfying |f(x)|<=|x|. Show that f is continuous at 0

    2. Relevant equations



    3. The attempt at a solution

    Really stuck on this cause I'm confused with the absolute values on this function.

    I *think* to show this you have to see if lim x>0+f(x) = lim x>0-f(x) = f(0) ?

    And I tried doing this:
    -|x|<=f(x)<=|x|
    lim x>0+|x|=0
    lim x>0- -|x|=0
    f(0)=|0|=0
    So they're all equal to 0.

    I don't know if this is right though...help?
     
  2. jcsd
  3. Jul 4, 2012 #2

    jedishrfu

    Staff: Mentor

    what is the definition of continuous? show that each part of the definition is satisfied by the function and you've shown it is continuous.

    If you notice that the slope of f(x) is bounded by the absolute value curve to be: 1>= slope >= -1 on either side of 0
     
    Last edited: Jul 4, 2012
  4. Jul 4, 2012 #3
    That's what I tried to do here.

    The definition of a function being continuous at 0 is that lim x>0 f(x) = f(0)
    So I tried to show the limit exists if the left and right hand limits are equal. And that this limit also equals f(0).

    I guess what I'm more confused about then, is the function itself. I don't know how to deal with the fact that the function is shown as part of an inequality, and with the absolute values
     
  5. Jul 4, 2012 #4
    Your work is nearly correct. You can now simply used the squeeze theorem, sandwiching f(x) between -|x| and |x|, so that it does satisfy the continuity definition by getting [itex]\lim_{x\to 0} f(x) = 0[/itex]
     
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