How is $ \lim_{z \to z_o} (z-z_o)\frac{f(z)}{g(z)} = \lim_{z \to z_o} \frac{f(z)}{g(z)-g(z_o)/(z-z_o)}= \frac{f(z_o)}{g'(z_o)}$? I was reading this proof in Gamelin Complex Analysis (page 196):

If $ f(z) $ and $ g(z) $ are analytic at $ z_o $ and if $ g(z) $ has a simple zero at $ z_o $

$$ Res[ \frac{f(z)}{g(z)},z_o ] = \frac{f(z_o)}{g'(z_o)} $$
In this case $ f(z)/g(z) $ has a simple pole at $ z_o$

$$  \lim_{z \to z_o} (z-z_o)\frac{f(z)}{g(z)} = \lim_{z \to z_o} \frac{f(z)}{g(z)-g(z_o)/(z-z_o)}= \frac{f(z_o)}{g'(z_o)}$$

I'm confused where does $g(z)-g(z_o)$ in the denominator come from?
Since $g(z)$ has a simple zero in $z_0$ you can freely add or subtract $g(z_0)$ in the denominator, because $g(z_0)=0$:
$$
(z-z_0) \frac{f(z)}{g(z)}=(z-z_0)\frac{f(z)}{g(z)-g(z_0)}=\frac{f(z)}{\frac{g(z)-g(z_0)}{z-z_0}}\to_{z\to z_0} \frac{f(z_0)}{g'(z_0)}.
$$
both numerator and denominator are of the form 0/0. So use L'Hospital rule.