Let $f,g$ both analythic in neighbourhood of $z_0$ and they both have zero of multiplicity $n$ in $z_0$. Prove that $\lim\limits_{z\to z_0}\frac{f(z)}{g(z)}=\lim\limits_{z\to z_0}\frac{f^\prime(z)}{g^\prime(z)}$. You mustn't use l'hospital rule but rather prove it in this private case.
I thought using the tecnique here but seems like it doesn't use the fact about the multiplicity of the zeros. I know that $\exists h(z),k(z)$ s.t $$f(z)=(z-z_0)^nk(z)\\ g(z)=(z-z_0)^nh(z)$$ but after substitution remain only $k(z),h(z)$ functions which are diffrent than $g(z),g(z)$. How can I prove this "private -case of l'hospital"?
Hint: First, prove the fact for $n=1$, then use induction.