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Limit proof (using delta-epsilon definition)

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f, g : A [tex]\rightarrow[/tex] R be functions and let c [tex]\in[/tex] R be a limit point of A. Assume that lim(x-> c): f(x) = L
    and lim(x->c): g(x) = M. Use the epsilon-delta definition to prove:
    lim(x->c) f(x)g(x)=LM


    2. Relevant equations



    3. The attempt at a solution
    alright in order to prove this, for epsilon>0, I need to find a delta>0 so that for 0<|x-c|<delta,
    |f(x)g(x)-LM|<epsilon.

    This is what I tried so far:

    |f(x)g(x)-LM|= |f(x)g(x)-Lg(x)+Lg(x)-LM|= |(f(x)-L)(g(x))+L(g(x)-M)|
    By the triangle inequality, that is less than or equal to:

    |(f(x)-L)(g(x)|+|L(g(x)-M)|= |f(x)-L||g(x)| + |L||g(x)-M|

    Well I know that since lim(x->c) f(x)=L, that means for epsilon>0 there exists a delta(1) so that for 0<|x-c|<delta(1)
    |f(x)-L|<epsilon

    and also since lim(x->c) g(x)=M, that means for epsilon>0 there exists a delta(2) so that for 0<|x-c|<delta(2)
    |g(x)-M|<epsilon

    For |f(x)-L||g(x)| + |L||g(x)-M|, I am trying to make each part at most equal to epsilon/2.

    This is where I get stuck, I don't know what to do with |f(x)-L||g(x)|, because g(x) isn't a constant.... any help would be great!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 1, 2009 #2

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    You may want to define epsilon1 and epsilon2 applied to the two separate limits. I think you'll have to again use the closeness of g(x) to M in the part where you are stuck so that epsilon1 (for f) is also a function of epsilon2 (for g). (Or maybe the other way around?)

    Beyond that you can try further alternative manipulation and application of the triangle inequality to see if you can isolate out that troublesome case.

    Hmmm....just trying some expressions...
    (f-L)(g-M) = fg -Lg -fM + MG.... (f-L)(g+M)/2 + (f+L)(g-M)/2 = fg -LM ...??? (f-L)(g-M + 2M)/2 < ?

    Something there might be useful or hint at something useful... or maybe not. I haven't gone through the proof myself yet.
     
  4. Nov 1, 2009 #3
    Let epsilon=1. Then there must exist a corresponding delta such that |g(x)-M|<1. How does this help?
     
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