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Derivative fraction proof

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data
    If [tex]\lim_{z\rightarrow z_0}f(z)=A[/tex] and [tex]\lim_{z\rightarrow z_0}g(z)=B[/tex] then prove that [tex]\lim_{z\rightarrow z_0}\frac{f(z)}{g(z)}=\frac{A}{B}[/tex] for [tex]B\neq0[/tex]

    3. The attempt at a solution

    I write [tex]f(z)=A+\epsilon_1(z)[/tex] and [tex]g(z)=B+\epsilon_2(z)[/tex], where the epsilon-functions tend to zero as z tends to z_0. I now write

    And since the above can be made arbitrarily small by letting z tend to z_0, I am done, or what do you think?
  2. jcsd
  3. Jun 16, 2009 #2


    Staff: Mentor

    I think you are not done. Given a positive number [itex]\epsilon[/itex], you have to demonstrate the existence of a positive number [itex]\delta[/itex] such that |x - x0| < [itex]\delta[/itex] implies that |f(x)/g(x) - A/B| < [itex]\epsilon[/itex].
  4. Jun 16, 2009 #3


    User Avatar
    Homework Helper

    Looks good to me. Check that [tex]\lim_{z\rightarrow z_0} f(z) = A[/tex] is equivalent to [tex]f(z) = A+\epsilon_1 (z)[/tex] where the epsilon-function goes to zero as [tex]z\rightarrow z_0[/tex], then you are in fact done.
  5. Jun 16, 2009 #4

    I have [tex]\epsilon_1(z)=f'(\xi)(z-z_0)[/tex] and [tex]\epsilon_2(z)=g'(\zeta)(z-z_0)[/tex] so

    \frac {|B\epsilon_1(z)|+|A\epsilon_2(z)|}{|B^2+B\epsilon _2(z)|}=\frac{(|Bf'(\xi)|+|Ag'(\zeta)|)|z-z_0|}{|B^2+Bg'(\zeta)(z-z_0)|}.

    Now if [tex]\delta=\frac{1}{k|g'(\zeta)|}[/tex] and [tex]|z-z_0|<\delta[/tex] and k>1 I get

    \frac{(|Bf'(\xi)|+|Ag'(\zeta)|)|z-z_0|}{|B^2+Bg'(\zeta)(z-z_0)|}\le \frac{1}{k}\frac{(|Bf'(\xi)|+|Ag'(\zeta)|)|g'(\zeta)|}{|B|^2-\frac{|B|}{k}}\le\frac{1}{k}C

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