TITLE: Is L'Hopitals rule applicable to complex functions?
QUESTION [49 upvotes]: I have a question about something I'm wondering about. I've read somewhere that
L'Hopitals rule can also be applied to complex functions, when they are analytic.
So if have for instance:
$$
\lim_{z \rightarrow 0} \frac{\log(1+z)}{z} \stackrel{?}{=} \lim_{z \rightarrow 0} \frac{1}{(1+z)} = 1
$$
Now i'm wondering if this is correct? Also if we take $|z|<1$, is it then correct?
Thanks,

REPLY [67 votes]: L'Hopital's rule is a local statement: it concerns  the behavior of functions near a particular point. The global issues (multivaluedness, branch cuts) are irrelevant. For example, if you consider $\lim_{z\to 0}$, then it's automatic that only small values of $z$ are in play.  Saying "take $|z|<1$" is redundant.
Generally, you have a point $a\in\mathbb C$ and some neighborhood of $a$ in which   $f,g$ are holomorphic. If $f(a)=g(a)=0$, then 
$$\lim_{z\to a}\frac{f(z)}{z-a}=f'(a),\qquad \lim_{z\to a}\frac{g(z)}{z-a}=g'(a) \tag{1}$$
hence 
$$\lim_{z\to a}\frac{f(z)}{g(z)}= \lim_{z\to a}\frac{f(z)/(z-a)}{g(z)/(z-a)} =\frac{f'(a)}{ g'(a)}$$
Note that the above is a  simple special case of  the L'Hopital's rule, because we have (1). It's basically just the definition of derivative.