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Proving continuity

  1. Oct 17, 2006 #1
    I am trying to fully understand this example from a textbook I am reading:

    http://img59.imageshack.us/img59/9237/continuityyn8.jpg [Broken]

    What I am not understanding is how they are proving it for [-1,1].. The way I see it is they proved that the function is continuous for all values in it's domain...

    For example, I thought up this problem on my own to help me understand :

    Given [tex] f(x)=1-\frac{1}{x-4} [/tex], prove that f(x) is continuous in the interval [-1,30] (Obviously it's not continuous at x=4.) The problem is that I don't see the connection between the interval and the solution...

    I can just go ahead and prove that [tex]\lim_{x\rightarrow a}f(x)=f(a)[/tex]... Which was stated in my text as meaning that the function is continuous... which it obviously isnt.

    [tex]\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}(1-\frac{1}{x-4})[/tex]
    [tex]=1-\lim_{x\rightarrow a}\frac{1}{x-4}[/tex]

    Can someone cure my confusion? Thanks guys.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 17, 2006 #2
    I am not sure as to whether or not the latex is really screwed up in my post, I edit it and it looks completely diff from what I see when I refresh the post. Click on the indivual latex box's to see what I meant to type if it comes up screwy. Maybe it's just my pc.

    Nevermind, it's fine now.
    Last edited: Oct 17, 2006
  4. Oct 17, 2006 #3
    oh my.. I see it now, limf(x)=f(a) for all values except a=4... since that would result in an undefined statement. But I still don't understand how they are testing for the interval [-1,1].... How would I test my made up example in the interval [-1,1]?
    Last edited: Oct 17, 2006
  5. Oct 17, 2006 #4


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    They are, indeed, showing it for the values in the function's PRESCRIBED DOMAIN, that is the interval from -1 to 1
  6. Oct 17, 2006 #5


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    The only time they implicity use the assmption that x was in [-1,1] was in moving the limit inside the square root. sqrt(x) is only continuous for x>=0, (namely because it is only defined here), so they needed that fact that 1-x^2 was non-negative.
  7. Oct 17, 2006 #6
    Okay I understand what they were doing now, thanks :)

    edit:Nevermind this other question, I reread the definition/example and it became clear.
    Last edited: Oct 17, 2006
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