Prove L' Hopitals Rule with Complex numbers Suppose that $f(z_{0}= g(z_{0}= 0$ and that both $f'(z_{0})$ and $g'(z_{0})$ exist with $g'(z_{0})\neq0$. Use the definition to prove that $\displaystyle \lim_{z\rightarrow z_{0}} \frac{f(z)}{g(z)}= \frac{f'(z)}{g'(z)}. $
My proof is as follows but i think it breaks down when i cross cancel the $z-z_0$ and the $z$ in the final step should be a $z_0$.
 
 A: The line that
\begin{align*}
\lim_{z\rightarrow z_{0}}\dfrac{f(z)}{g(z)}=\dfrac{f'(z)}{g'(z)}
\end{align*}
is not valid.
Rather, one uses holomorphicity to claim that
\begin{align*}
\lim_{z\rightarrow z_{0}}\dfrac{f(z)}{g(z)}&=\lim_{z\rightarrow z_{0}}\dfrac{f(z_{0})+f'(z_{0})(z-z_{0})+\dfrac{1}{2}f''(z_{0})(z-z_{0})^{2}+\cdots}{g(z_{0})+g'(z_{0})(z-z_{0})+\dfrac{1}{2}g''(z_{0})(z-z_{0})^{2}+\cdots}\\
&=\lim_{z\rightarrow z_{0}}\dfrac{f'(z_{0})+\dfrac{1}{2}f''(z_{0})(z-z_{0})+\cdots}{g'(z_{0})+\dfrac{1}{2}g''(z_{0})(z-z_{0})+\cdots}\\
&=\dfrac{f'(z_{0})}{g'(z_{0})}.
\end{align*}
A: Your proof is not valid but we can make a proof as you say.
Suppose that $f(x_0), g(x_0)$ are both $0$ and $f'(x_0), g'(x_0)$ exist, $g'(x_0)\neq 0$ 
Then we have:
$$\lim_{z \to z_0} \frac{f(x)}{g(x)}=\lim_{z \to z_0} \frac{f(x)}{g(x)}\left(  \frac{z-z_0}{z-z_0}\right)$$
$$=\lim_{z \to z_0} \frac{f(x)-f(x_0)}{g(x)-g(x_0)}\left( \frac{z-z_0}{z-z_0}\right)$$
$$=\lim_{z \to z_0} \frac{f(x)-f(x_0)}{z-z_0}\left(  \frac{z-z_0}{g(x)-g(x_0)}\right)$$
$$=\lim_{z \to z_0} \frac{f(x)-f(x_0)}{z-z_0} *\lim_{z \to z_0}\frac{z-z_0}{g(x)-g(x_0)}$$
$$=\lim_{z \to z_0} \frac{f(x)-f(x_0)}{z-z_0} \left( \lim_{z \to z_0}\frac{g(x)-g(x_0)}{z-z_0}\right)^{-1}$$
$$=\frac{f'(x_0)}{g'(x_0)}$$
