Find $\lim _ { z \rightarrow z _ { 0 } } \frac { f ( z ) } { g ( z ) }$ with f and g holomorphic and $z_0$ a zero of $f$ and $g$

Question

Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$

My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have : $$ \begin{aligned} f ( z ) & = f \left( z _ { 0 } \right) + f ^ { \prime } \left( z _ { 0 } \right) \left( z - z _ { 0 } \right) + \frac { f ^ { \prime \prime } \left( z _ { 0 } \right) } { 2 ! } \left( z - z _ { 0 } \right) ^ { 2 } + \cdots \\ & = \left( z - z _ { 0 } \right) ^ { l } \left[ \frac { f ^ { ( l ) } \left( z _ { 0 } \right) } { l ! } + \sum _ { n = l + 1 } ^ { + \infty } \frac { f ^ { ( n ) } \left( z _ { 0 } \right) } { n ! } \left( z - z _ { 0 } \right) ^ { n - l } \right] \\ & = \left( z - z _ { 0 } \right) ^ { l } \frac { f ^ { ( l ) } \left( z _ { 0 } \right) } { l ! } + \left( z - z _ { 0 } \right) ^ { l } F ( z ) \end{aligned} $$

In the same way we obtain : $$ \begin{aligned} g ( z ) & = \left( z - z _ { 0 } \right) ^ { l } \left[ \frac { g ^ { ( l ) } \left( z _ { 0 } \right) } { l ! } + \sum _ { n = l + 1 } ^ { + \infty } \frac { g ^ { ( n ) } \left( z _ { 0 } \right) } { n ! } \left( z - z _ { 0 } \right) ^ { n - l } \right] \\ & = \left( z - z _ { 0 } \right) ^ { l } \frac { g ^ { ( l ) } \left( z _ { 0 } \right) } { l ! } + \left( z - z _ { 0 } \right) ^ { l } G ( z ) \end{aligned} $$

My issue : To conclude I would like to write the following steps : $$ \lim _ { z \rightarrow z _ { 0 } } \left[ \frac { f ( z ) } { g ( z ) } \right] = \lim _ { z \rightarrow z _ { 0 } } \left[ \frac { \frac { f ^ { ( l ) } \left( z _ { 0 } \right) } { l ! } + F ( z ) } { \frac { g ^ { ( l ) } \left( z _ { 0 } \right) } { l ! } + G ( z ) } \right] = \frac { f ^ { ( l ) } \left( z _ { 0 } \right) } { g ^ { ( l ) } \left( z _ { 0 } \right) } $$

1. Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } \left( z _ { 0 } \right) } \neq 0$.
2. $G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?

Answer 1

A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) \neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) \neq 0$. This immediately gives the result since $\frac {f(z)} {g(z)} \to 0$ if $l >k$ and $\frac {f(z)} {g(z)} \to \frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?