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Proving division of continuous functions

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Okay, so if f and g are continuous functions at a, then prove that f/g is continuous at a if and only if g(a) # 0

    2. Relevant equations
    Assuming to start off the g(a)#0, by the delta-epsilon definition of continuity, basically, We know that |f(x)| and |g(x)| are bounded.

    3. The attempt at a solution

    I have messed around with the end result that we need, which is when |x-a|< delta
    |f(x)/g(x)-f(a)g(a)|<Epsilon. This is what I've come up with:

    |1/(g(x)g(a))|*|f(x)g(a)-f(a)g(x)|.
    By looking at each piece it seems like they can be bounded as well. However, how do i manipulate what each one is bounded by so that when I multiply and add everything out, I get a nice simple Epsilon?
     
  2. jcsd
  3. Nov 26, 2007 #2
    you know...I could be going about it completely the wrong way too..
     
  4. Nov 26, 2007 #3

    Hurkyl

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    You've already proven a bunch of limit theorems. Can you use those limit theorems to derive this one?


    P.S. I assume by # you meant [itex]\neq[/itex]; you should say that someplace rather than just using it without warning.
     
  5. Nov 26, 2007 #4

    Hurkyl

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    If both pieces are bounded by [itex]\sqrt{\epsilon}[/itex], then what is their product bounded by?

    (I assume by "piece" you meant the two multiplicands)
     
  6. Nov 26, 2007 #5
    sorry about that, I didn't realize you could do that...I'm new here,
    And I need to use the rigorous definitions of continuity, so deriving it here from what I know about the bunches of limit theorems I know isn't sufficient.
     
  7. Nov 26, 2007 #6
    It doesn't seem that easy. Maybe I should back up.
    The fact that f and g are continuous at a mean that for |x-a|<delta,
    |f(x)-f(a)|<Something involving epsilon(1)
    |g(x)-g(a)|<something involving ep(2)

    By triangle inequality,
    |f(x)|< something involving E(1) + |f(a)|
    |g(x)|< something involving E(2) + |g(a)|

    Since this is what we know, we have to start multiplying around and around using these two inequalities to get the final result |1/(g(x)g(a))|*|f(x)g(a)-f(a)g(x)|. For example,
    |f(x)g(a)|< something involving E(1)*|g(a)|+|f(a)g(a)|.

    Now I'm confusing myself again, Am I making this too complicated? or what
     
  8. Nov 26, 2007 #7

    Hurkyl

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    There's a famous trick that's very useful for manipulating absolute value inequalities:

    AB - CD = AB - AD + AD - CD = A(B-D) + (A-C)D


    So if B-D and A-C are things you happen to know something about...
     
  9. Nov 26, 2007 #8
    that only seems useful if I'm trying to prove that f*g is continuous at a, but I'm lost as to see why that's useful for f/g
     
  10. Nov 27, 2007 #9

    Office_Shredder

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    Because you have

    |1/(g(x)g(a))|*|f(x)g(a)-f(a)g(x)|

    Specifically, the
    |f(x)g(a)-f(a)g(x)|

    term can be split. to deal with the 1/g(x), note that eventually g(x)>g(a)-e (e being epsilon) for all x within delta of a, for any epsilon
     
  11. Nov 28, 2007 #10
    Okay, abandon my initial approach. I think it'd be ten times easier if I show that 1/g(x) is continuous at a. Since g(a) is not zero, then g(x) has to bounded away from zero. Which means that 1/g(x) will not get infinitely large, but rather, will be bounded by something else. Now if only I could put that into a nice algebraic argument that would have a bit more clout.
    If |g(x)-g(a)|< e then how do you get that g(x) > g(a) - e without ignoring the absolute values?
     
  12. Nov 28, 2007 #11

    Office_Shredder

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    By definition, that means -e< g(x)-g(a) < e so g(a)-e<g(x) and g(x)< g(a)+e

    Geometric intuition time: If |x-a|<e for any kind of x, a and e, that means that x is within e of a.
     
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