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Proof of l'hopital 3

  1. Apr 23, 2005 #1
    if lim f(x)= infinity= lim g(x)
    x->infinity x->infinty

    and lim f'(x)/g'(x)=infinity
    x-> infinity

    then lim f(x)/g(x)=inifity
    x-> inifinity

    The above fact is what im trying to prove. From my notes, i see the following:

    For m>0, choose k>0, such that if x> k* and g(x),f(x)>0,

    then f'(x)/g'(x)> m(4/3).


    this is actually where i get lost (so early into the process). can someone explain to me where exactly the prof is headed to with this info? also, is k a functional value through m? if so...how do i choose such k?
     
  2. jcsd
  3. Apr 23, 2005 #2
    i just realized that my post is quite confusing. i suppose im not too sure what to ask. If you dont understand my original post, perhaps u can just help me start off the proof. thanks.
     
  4. Apr 23, 2005 #3
    i also have the following hints to use after the hints in the original post:

    fix a>k, then by cauchy mvt,

    f(x)-f(a) f'(c)
    _______ = ____ > m(4/3) for x>a>k*, and such that
    g(x)-g(a) g'(c)

    g(x)>g(a)>0 and f(x)> f(a)> 0 where c>a>k*
     
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