# Something that looks like L'Hopital for the complex case

If we assume that $\displaystyle f,g:{B_R}\left( a \right) \to C$ are homomorphic, and $\displaystyle f\left( a \right) = {f^{\left( 1 \right)}}\left( a \right) = ... = {f^{\left( {n - 1} \right)}}\left( a \right) = 0$, $\displaystyle g\left( a \right) = {g^{\left( 1 \right)}}\left( a \right) = ... = {g^{\left( {n - 1} \right)}}\left( a \right) = 0$ with $\displaystyle {g^{\left( n \right)}}\left( a \right) \ne 0$, then show that $\displaystyle \mathop {\lim }\limits_{z \to a} \frac{{f\left( z \right)}}{{g\left( z \right)}} = \frac{{{f^{\left( n \right)}}\left( z \right)}}{{{g^{\left( n \right)}}\left( z \right)}}$.
I can show that $\displaystyle \frac{{{f^{\left( n \right)}}\left( z \right)}}{{{g^{\left( n \right)}}\left( z \right)}}$ exists quite trivially, but I'm not sure how I can say anything about how the two quotients are related, and how the limit can be said to equal the RHS. The question looks similar to L'Hopital, but that is only defined for the real case, whereas these are functions defined on the complex plane.