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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,

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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

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