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Prove limit theorem using epsilon-delta

  1. Sep 24, 2011 #1
    Hey i am trying to understand Spivak's proof of lim x->a of f(x)g(x)=lm (where l is limit of f(x) and m is lim of g(x) )..but i think he is skipping many steps and at one point i dont understand why he is doing something..

    ok so the following i understand:

    [itex]\left|f(x)g(x)-lm\right|[/itex]< E
    [itex]\leq[/itex][itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E


    [itex]\left|g(x)-m\right|[/itex]< [itex]\frac{E}{2(\left|l\right|+1)}[/itex]



    ok so now i am guessing that we wanna say that [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2 so that E/2 + E/2 = E
    so can u just say:

    cus i mean m is just a constant..it cant be restricted like f(x) was...

    [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<[itex]\left|m\right|[/itex](E/(2[itex]\left|m\right|[/itex]))= E/2

    so then:
    [itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]< E/2 + E/2 =E

    when [itex]\left|f(x)-l\right|[/itex] < min (1, E/(2|m|) ) and [itex]\left|g(x)-m\right|[/itex]<[itex]\frac{E}{2(\left|l\right|+1)}[/itex]

    but Spivak is saying that [itex]\left|f(x)-l\right|[/itex] < min (1, E/( 2(|m|+1) ) ) and i have no clue why... help plsss?
  2. jcsd
  3. Sep 24, 2011 #2
    Yes, but now you have done something like


    so you have two delta's. But you only want one delta. So if we take [itex]\delta=\min(\delta_1,\delta_2)[/itex], then [itex]|f(x)-l|<\delta[/itex] implies that [itex]|f(x)-l|<\delta_1[/itex]. And the same with g(x).

    So Spivak has two delta's:


    and he combines it to one delta. This is what he does.
  4. Sep 24, 2011 #3
    I know that you have to combine the two deltas by taking the minimum delta, but my question is, how does Spivak get that [itex]\left|f(x)-l\right|[/itex] < min (1, E/( 2(|m|+1) ) ) ?

    He doesnt get [itex]\left|f(x)-l\right|[/itex] < min (1, E/( 2(|m|) )

    He is adding some random 1 next to the |m| ?
  5. Sep 24, 2011 #4
    He's combining the delta's



  6. Sep 24, 2011 #5
    But delta#2 is not [itex]\frac{1}{2|m|+1}[/itex]

    it is [itex]\frac{1}{2(|l|+1)}[/itex]
  7. Sep 24, 2011 #6
    Ah yes, it seems you're right then. I suspect Spivak adds the +1 to make sure that he didn't divide by 0. That is: if l=0, then you would have 1/0 if you didn't add +1.
  8. Sep 24, 2011 #7
    haha its legal to just do that? i mean this is supposed to be a rigid proof...i dont think u can just add a random 1 without explaining why?

    why not just say that m[itex]\neq0[/itex], l[itex]\neq0[/itex] ???

    i think the 1 comes from a more logical step?
  9. Sep 24, 2011 #8
    It is in this case, since


    So you're only making the delta smaller. This is certainly allowed.

    This is also ok, but then you would have to give a different proof in the case that l=0 or m=0.
  10. Sep 24, 2011 #9
    ohhh gotcha thanks....are you sure though that this 1 cant possibly follow from some earlier step?? i just want to make sure because it is bothering me lol
  11. Sep 24, 2011 #10
    I looked at Spivak, and this is the only possible reason I see why he would want to add +1.
  12. Sep 25, 2011 #11
    You should try to see why Spivak's proof works but when I learned this from Spivak it all felt very mysterious. See if you can do the proof yourself:

    | (a+[itex]\delta[/itex])(b+[itex]\delta[/itex]) - ab | < [itex]\epsilon[/itex]

    | (a+b)[itex]\delta[/itex] + [itex]\delta[/itex]2| < [itex]\epsilon[/itex]

    can you see why its not too hard to find a [itex]\delta[/itex] that works.


    I'm sorry if this post doesn't help but I never understood this proof unless I could find it out myself.
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