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Prove limit theorem f(x)g(x)=lm

  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?
    Last edited: Sep 24, 2011
  2. jcsd
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