How can the limit of a derivative fraction be proven for non-zero values?

[daudaudaudau]

Homework Statement

If $$\lim_{z\rightarrow z_0}f(z)=A$$ and $$\lim_{z\rightarrow z_0}g(z)=B$$ then prove that $$\lim_{z\rightarrow z_0}\frac{f(z)}{g(z)}=\frac{A}{B}$$ for $$B\neq0$$

The Attempt at a Solution

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

$$\left|\frac{f(z)}{g(z)}-\frac{A}{B}\right|=\left|\frac{A+\epsilon_1(z)}{B+\epsilon_2(z)}-\frac{A}{B}\right|=\left|\frac{AB+B\epsilon_1(z)-AB-A\epsilon_2(z)}{B^2+B\epsilon_2(z)}\right|\le\frac{|B\epsilon_1(z)|+|A\epsilon_2(z)|}{|B^2+B\epsilon_2(z)|}$$
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 $\epsilon$, you have to demonstrate the existence of a positive number $\delta$ such that |x - x₀| < $\delta$ implies that |f(x)/g(x) - A/B| < $\epsilon$.

 

[benorin]

Looks good to me. Check that $$\lim_{z\rightarrow z_0} f(z) = A$$ is equivalent to $$f(z) = A+\epsilon_1 (z)$$ where the epsilon-function goes to zero as $$z\rightarrow z_0$$, then you are in fact done.

 

[daudaudaudau]

Mark44:

I have $$\epsilon_1(z)=f'(\xi)(z-z_0)$$ and $$\epsilon_2(z)=g'(\zeta)(z-z_0)$$ 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 $$\delta=\frac{1}{k|g'(\zeta)|}$$ and $$|z-z_0|<\delta$$ and k>1 I get

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

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