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?


  • $\begingroup$ Your limit is just the derivative of $log(1+z)$ at $0$, and it is 1. What do you mean by taking $|z| < 1$? I think that we just care about a neighborhood of $0$. $\endgroup$ – Du Phan Dec 11 '13 at 11:37
  • $\begingroup$ I heard that L'Hopitals rule can only be applied to analytic functions in a certain region. So if we have that $|z|<1$, then we get no problems with the definition of the $\log$ right? $\endgroup$ – user112167 Dec 11 '13 at 11:40
  • 3
    $\begingroup$ By the way, if $f, g$ analytic at $a$, and $f(a) = g(a) = 0$, then $\lim_{z\to a} \frac{f(z)}{g(z)} = \lim_{z\to a} \frac{f(z) - f(a)}{z-a}.\frac{z-a}{g(z)-g(a)} = \frac{f'(a)}{g'(a)}$. $\endgroup$ – Du Phan Dec 11 '13 at 11:41
  • $\begingroup$ Since you're interested in the limit at $0$, bounding the functions to the unit circle is irrelevant. $\endgroup$ – egreg Dec 11 '13 at 11:41
  • 1
    $\begingroup$ Both $\log(1+z)$ and $1/z$ are meromorphic at 0. For meromorphic functions the Laurent series of their product is the product of their Laurent series: so you have $$ z^{-1}\log(1+z) = z^{-1}(0 + z + O(z^2)) = 1 + O(z) $$ and is holomorphic. $\endgroup$ – Willie Wong Dec 11 '13 at 13:45

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.

| cite | improve this answer | |
  • $\begingroup$ We can use this when the limit goes to infinity, right ? $\endgroup$ – onurcanbektas Nov 12 '18 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.