Derivative fraction proof

daudaudaudau

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

[tex]
\left|\frac{f(z)}{g(z)}-\frac{A}{B}\right|=\left|\frac{A+\epsilon_1(z)}{B+\epsilon_2(z)}-\frac{A}{B}\right|=\left|\frac{AB+B\epsilon_1(z)-AB-A\epsilon_2(z)}{B^2+B\epsilon_2(z)}\right|\le\frac{|B\epsilon_1(z)|+|A\ epsilon_2(z)|}{|B^2+B\epsilon_2(z)|}
[/tex]
And since the above can be made arbitrarily small by letting z tend to z_0, I am done, or what do you think?

Mark44

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].

benorin

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.

daudaudaudau

Mark44:

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

[tex]
\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)|}.
[/tex]

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

[tex]
\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
[/tex]

where
[tex]
C=\frac{(|Bf'(\xi)|+|Ag'(\zeta)|)|g'(\zeta)|}{|B|^2-|B|}
[/tex]